May 28, 2026
We study Lusin-measurable functions with values in locally convex spaces. In particular, the behavior of pointwise limits of sequences of Lusin-measurable functions and exhibit pathological phenomena arising in the nonmetrizable setting. Moreover, we establish approximation and density results for \(L^p\)-spaces constructed with this notion of measurability, including the density of simple functions in Hausdorff locally convex spaces and convergence results obtained through dyadic approximations.
On \(L^p\)-spaces of functions with values in locally convex spaces
Matthieu F. Pinaud1, Humberto Prado
MSC 2020 subject classification: 46G10 (primary); 28A20, 28B05, 46A03
Keywords: Locally convex spaces, vector-valued functions, \(L^p\)-spaces, Lusin-measurable functions
Let \(a<b\) be real numbers. We consider the closed interval \([a,b]\) endowed with its Borel \(\sigma\)-algebra \(\mathcal{B}([a,b])\) and \(\lambda\) is the Lebesgue measure. For a real Banach space \(E\), the classical theory of Lebesgue spaces \(L^p([a,b],E)\) of vector-valued functions is well understood via Bochner-measurability.
Many problems in infinite-dimensional analysis naturally arise in the setting of locally convex spaces. For example, the spaces encountered in distribution theory are, in general, non-metrizable locally convex spaces. Furthermore, solution curves of
quantum stochastic differential equations take values in spaces of continuous linear operators between nuclear locally convex spaces (see, e.g., [1]). Likewise, the
theory of regularity of Lie groups (and half-Lie groups) is concerned with \(L^p\) functions taking values in sequentially complete locally convex Lie algebras. (see e.g. [2]–[6]).
Let \(E\) be a real locally convex space and \(E'\) its topological dual space. Then we recall different notions of measurability for a functions \(\gamma:[a,b]\to E\).
The function \(\gamma\) is Borel-measurable if \(\gamma^{-1}(A) \in \mathcal{B}([a,b])\) for each \(A\in \mathcal{B}(E)\).
The function \(\gamma\) is Lusin-measurable if for each \(\varepsilon>0\) there exists a compact subset \(K_\varepsilon\) of \([a,b]\) such that \(\gamma|_{K_\varepsilon}\) is continuous and \(\lambda\left ([a,b]\setminus K_\varepsilon\right)\leq \varepsilon\).
The function \(\gamma\) is strongly-measurable if it is the pointwise limit a.e. of a sequence of simple functions \((\beta_n)_n\).
The function \(\gamma\) is weakly-measurable if for each \(f\in E'\), the function \(f\circ \gamma:[a,b]\to \mathbb{R}\) is Borel-measurable.
In the classical case, that is when \(E\) is finite-dimensional. Then all four notion of measurable functions, are equivalent. If \(E\) is any second countable topological vector space,
by the Lusin’s Theorem, Borel-measurability implies Lusin-measurability (see e.g. [7], [8]). If \(E\) is a Banach space, by the Pettis Measurability Theorem (see [7]), a function is strongly-measurable if and only if it is separably
valued and weakly-measurable. We notice that, whenever \(E\) is a not-metrizable locally convex space, then all these notion are not equivalents since locally convex spaces introduce several pathologies that do not arise in
the Banach setting. For instance, the sum of two Borel-measurable functions may fail to be a Borel-measurable function (see e.g. [9]) and the pointwise limit of
continuous functions may fail to be Lusin-measurable (see Remark 1).
The locally convex setting has already been studied by several authors (see e.g. [2], [3], [5], [10]–[12]). More recently, Natalie Nikitin has proved that, for every Lusin-measurable function \(\gamma\) there exists a Borel-measurable function \(\beta\) such that \(\gamma=\beta\) a.e.
In the case of Banach spaces, the classical theory of \(L^p\)-spaces is well understood. For example, for Fréchet spaces, in [2], H.
Glöckner has established classical results for \(L^p\)-spaces using Borel-measurability. In the locally convex setting, several authors have studied the construction of \(L^p\)-spaces using
Lusin-measurability (see e.g. [5], [6], [10]–[12]).
The article has been divided in two parts. The first one focuses on the study of the pointwise limit of a sequence of Lusin-measurable functions \((\gamma_n)_n\) and the conditions under which the pointwise limit is
Lusin-measurable.
In the second part, for \(1\leq p< \infty\), we focus on the \(L^p\)-space of Lusin-measurable functions. We show some classical density and convergence results. We exhibit a pathology
where the convergence in \(L^p\) does not imply any kind of pointwise convergence, in opposition to the classical theory. Moreover, we show that every \(L^p\)-function is the limit of a
sequence of simple function defined by dyadic partition.
Hereafter, we assume that all locally convex spaces are real vector spaces and are not necessarily Hausdorff. For details on locally convex spaces, we recall references [4], [13].
We recall the real-valued function \(g:[0,1]\to [a,b]\), \(g(t)=a+t(b-a)\). Now, if \(\gamma:[a,b]\to E\) is Lusin-measurable, then the function \(\gamma\circ g:[0,1]\to E\) is Lusin-measurable (see, [5]). This enables us to work with the closed interval \([0,1]\).
Let \(E\) and \(F\) be locally convex spaces. It is straightforward that each continuous functions is Lusin-measurable. If \(\gamma,\eta:[0,1]\to E\) are
Lusin-measurable functions and \(\varphi:E\to F\) is a continuous map, then \(\varphi\circ \gamma:[0,1]\to F\) is Lusin-measurable. Since \(E\) is a
topological vector space, then for each \(\alpha\in \mathbb{R}\), the function \[\alpha\gamma+\eta:[0,1]\to E,\quad t\mapsto \alpha\gamma(t)+\eta(t)\] is Lusin-measurable.
Moreover, if \(\gamma:[0,1]\to E\) is Lusin-measurable, then it is weakly-measurable. Indeed, for each \(\varepsilon>0\) and \(f\in E'\), there exists
a compact subset \(K_\varepsilon\) of \([0,1]\) such that \(\lambda([0,1]\setminus K_\varepsilon)<\varepsilon\) and \((f\circ\gamma)|_{K_\varepsilon} = f\circ (\gamma|_{K_\varepsilon})\) is continuous. Then \(f\circ \gamma:[0,1]\to \mathbb{R}\) is Lusin-measurable and hence is Borel-measurable.
Lemma 1. Let \(E\) be a locally convex space, \(A\in\mathcal{B}([0,1])\) and \(y_0\in E\). If \(\chi_A\) denotes the characteristic function of \(A\), then the function \[\beta:[0,1]\to E,\quad t\mapsto y_0\chi_A(t)\] is Lusin-measurable.
Proof. For \(\varepsilon>0\), we must obtain a compact set \(K_\varepsilon\subseteq [0,1]\) such that \(\lambda\left( [0,1]\setminus
K_\varepsilon\right)<\varepsilon\) and \[\beta|_{K_\varepsilon}:K_\varepsilon\to E,\quad t\mapsto \beta|_{K_\varepsilon}(t)= \left\{\begin{array}{ll} y_0, & \text{ if t\in A\cap K_\varepsilon} \\ 0, & \text{ if
t\in K_\varepsilon\setminus A}, \end{array}\right.\] Semiregularity is continuous, i.e., the set \(A\cap K_\varepsilon\) is open and closed in \(K_\varepsilon\).
By regularity of the Lebesgue measure, for \(\varepsilon>0\), there exists a compact subset \(F_1\) of \([0,1]\) with \(F_1\subseteq A\) such that \[\lambda(A\setminus F_1) < \varepsilon/2.\] and a compact subset \(F_2\) of \([0,1]\) with \(F_2\subseteq [0,1] \setminus A\) such that \[\lambda\left( \left( [0,1]\setminus A\right)\setminus F_2 \right) < \varepsilon/2.\] Next, we define \(K_\varepsilon=F_1\cup F_2\). Thus \[\begin{align} [0,1]\setminus K_\varepsilon & \subseteq [A\setminus F_1]\cup [ ([0,1]\setminus A)\setminus F_2].
\end{align}\] Hence \[\lambda([0,1]\setminus K_\varepsilon)\leq \lambda(A\setminus F_1)+\lambda(([0,1]\setminus A)\setminus F_2) < \varepsilon.\] Since \(F_1\) and \(F_2\) are closed subsets of \(K_\varepsilon\), the set \(A\cap K_\varepsilon\) is closed in \(K_\varepsilon\). We notice that
\[A\cap K_\varepsilon = F_1.\] Then \(F_1\) is open in \(K_\varepsilon\), since \[K_\varepsilon \setminus (A\cap K_\varepsilon) =
K_\varepsilon \cap F_1^c = F_2.\] Therefore, \(\beta|_{K_\varepsilon}\) is continuous. ◻
Definition 1. A function \(\beta:[0,1]\to E\) is called a simple functions if there exists \(y_1,...,y_n\in E\) and \(A_1,...,A_n\in \mathcal{B}([0,1])\) such that \[\beta=\sum_{i=1}^n y_i\chi_{A_i}.\] By the previous lemma, each simple function is Lusin-measurable. We denote the vector space of simple function by \(\mathcal{S}([0,1],E)\).
Lemma 2. Assume that \(E\) is a locally convex space and \(\gamma:[0,1]\to E\) a Lusin-measurable function. Then, for each \(\varepsilon>0\) and each continuous seminorm \(q\), there exists a compact subset \(K_{\varepsilon}\) of \([0,1]\) and a simple function \(\beta_{\varepsilon,q}:[0,1]\to E\) such that \(\lambda([0,1]\setminus K_{\varepsilon})<\varepsilon\), the function \(\gamma|_{K_\varepsilon}\) is continuous and \[\sup_{t\in K_\varepsilon} q(\gamma(t)-\beta_{\varepsilon,q}(t))< \varepsilon.\]
Proof. Let \(q\) be a continuous seminorm on \(E\) and \(\varepsilon>0\). Then there exists a compact subset \(K_\varepsilon\) such that \(\lambda([0,1]\setminus K_\varepsilon)<\varepsilon\) and \(\gamma|_{K_\varepsilon}\) is continuous. By compactness of \(K_\varepsilon\), the function \(\gamma|_{K_\varepsilon}\) is uniformly continuous with respect to \(q\), i.e., there exists \(\delta_{\varepsilon,q}>0\) such that \[\lvert t-s\lvert < \delta_{\varepsilon,q} \implies q(\gamma|_{K_\varepsilon}(t)-\gamma|_{K_\varepsilon}(s))<\varepsilon.\] Let \(\{t_0,...,t_n\}\) be a partition of \([0,1]\) such that \[\max_{1\leq i \leq n} \lvert t_i-t_{i-1}\lvert <\delta_{\varepsilon,q}.\] For \(i\in \{1,...,n\}\), if \([t_{i-1},t_i)\cap K_\varepsilon\neq \emptyset\), let us consider any \(s_i\in [t_{i-1},t_i)\cap K_\varepsilon\) and define the function
\(\beta_{\varepsilon,q}:[0,1]\to E\) by \[\beta_{\varepsilon,q}(t) = \sum\limits_{\substack{1\leq i\leq n \\ [t_{i-1},t_i)\cap K_\varepsilon\neq \emptyset}}
\gamma|_{K_\varepsilon}(s_i)\chi_{[t_{i-1},t_i)\cap K_\varepsilon}(t) + \gamma(1)\chi_{\{1\}}(t),\quad \forall t\in [0,1].\] Then \(\beta_{\varepsilon,q}\) is a simple function with each \([t_{i-1},t_i]\cap K_\varepsilon\) in \(\mathcal{B}([0,1])\).
Moreover, for each \(t\in K_\varepsilon\) there exists some \(j\in \{1,...,n\}\) such that \(t,s_j\in [t_{j-1},t_j]\), hence \(\lvert t-s_j\lvert < \delta_{\varepsilon,q}\) and \[q(\gamma|_{K_\varepsilon}(t)-\beta_{\varepsilon,q}(t)) = q(\gamma(t)- \gamma|_{K_\varepsilon}(s_j)) < \varepsilon.\] and \(\gamma(1)-\beta_{\varepsilon,q}(1)=0\). Therefore \[\sup_{t\in K_\varepsilon} q(\gamma|_{K_\varepsilon}(t)-\beta_{\varepsilon,q}(t))< \varepsilon.\] ◻
Remark 1. Let \(E\) be a locally convex space. For \(n\in \mathbb{N}\), let \(\gamma_n:[0,1]\to E\) be a sequence of Lusin-measurable functions. We denote the pointwise limit of this sequence by \[\gamma:[0,1]\to E,\quad \gamma(t) = \lim_{n\to \infty} \gamma_n(t).\] Then, we observe that even though each \(\gamma_n\) is continuous, the pointwise limit may not be Lusin-measurable. Indeed, let us consider the real vector space \[E=\{f:\mathbb{R}\to \mathbb{R}\},\] and for each element \(x_0\in \mathbb{R}\), we define the seminorm \[q_{x_0} (f) = \lvert f(x_0)\lvert ,\quad \forall f\in E.\] Then we endow the space \(E\) with the locally convex Hausdorff topology generated by the family of separating seminorms \(S=\{q_{x_0} : x_0\in \mathbb{R}\}\). Moreover, for each \(n\in \mathbb{N}\) we define \[\gamma_{n}:[0,1]\to E,\quad t\mapsto \gamma_{n}(t).\] Where \[\begin{align} \gamma_{n}(t)(x) &=\left\{ \begin{array}{ll} 1-n\lvert x-t \lvert & \text{, if } x\in \left[t-1/n, t+1/n\right] \\ 0 & \text{, if } x\not\in \left[t-1/n, t+1/n\right] \end{array}\right.\\ &= \max \{ 1-n\lvert x-t\lvert, 0\}, \end{align}\] for each \(x\in \mathbb{R}.\) It is easy to see that \((\gamma_n)_n\) is a sequence in \(C([0,1],E)\). Indeed, for each \(n\in \mathbb{N}\) and \(x\in \mathbb{R}\) the map \[\phi_{n,x}:[0,1]\to \mathbb{R},\quad \phi_{n,x}(t)=\max \{ 1-n\lvert x-t\lvert, 0\}\] is continuous. We identify \(E\) with \(\mathbb{R}^{\mathbb{R}}\) with the product topology, and for each \(x_0\in \mathbb{R}\) we denote the projection \[\pi_{x_0}: \prod_{x\in \mathbb{R}} \mathbb{R}\to \mathbb{R},\quad (y_x)_{x\in\mathbb{R}}\mapsto y_{x_0}.\] Then for all \(x_0\in\mathbb{R}\) and \(t\in[0,1]\) we have that the function \[\pi_{x_0}\circ \gamma_n:[0,1]\to \mathbb{R},\quad t\mapsto \pi_{x_0}\circ \gamma_n(t) = \phi_{n,x_0}(t)\] is continuous. Hence each \(\gamma_n:[0,1]\to E\) is continuous and consequently, Lusin-measurable. This sequence converges pointwise to \[\gamma:[0,1]\to E,\quad t\mapsto \gamma(t).\] Where \[\begin{align} \gamma(t)(x) &= \left\{\begin{array}{ll} 1 & \text{, if } x=t \\ 0 & \text{, if } x\neq t, \end{array}\right. \end{align}\] for each \(x\in \mathbb{R}.\) Then \(\gamma\) is not Lusin-measurable. Indeed, let \(K\subseteq [0,1]\) be a compact subset such that \(\gamma|_K\) is continuous. Thus, by the the definition of the product topology, the map \[\phi_x:K\to \mathbb{R},\quad s\mapsto \gamma(s)(x)=\delta_s (x)\] is continuous. Moreover, if \(x\not\in K\) then \(\phi_x=0\) and if \(x\in K\) the set \[\phi_x^{-1}\left(\left(\frac{1}{2},\infty\right)\right) = \{ x\}\] is open in \(K\). Now, since \(K\) is equipped with the discrete topology, compactness implies that \(K\) is finite. Therefore, \(\lambda(K)=0\) and \[\lambda([0,1]\setminus K ) = 1.\] Thus, \(\gamma\) can not be Lusin-measurable. However, \(\gamma\) is Borel-measurable. Indeed, let \(x_0\in \mathbb{R}\). Then \[\pi_{x_0}\circ \gamma:[0,1]\to \mathbb{R},\quad t\mapsto \delta_t(x_0) = \chi_{\{x_0\}}(t).\] Since each singleton \(\{x_0\}\) is measurable in \(\mathbb{R}\), for all \(x\in \mathbb{R}\), the coordinate function \(\pi_{x}\circ\gamma\) is Borel-measurable. Let \(U\subseteq E\cong\mathbb{R}^{\mathbb{R}}\) be a basic open set in the product topology. Then there exists \(x_1,...,x_k\in \mathbb{R}\) and open subsets \(U_1,...,U_k\subseteq \mathbb{R}\) such that \[U = \bigcap_{i=1}^k \pi_{x_i}^{-1}(U_i).\] Therefore, \[\gamma^{-1}(U) = \bigcap_{i=1}^k (\pi_{x_i}\circ \gamma)^{-1}(U_i)= \bigcap_{i=1}^k \chi_{\{x_i\}}^{-1}(U_i)\] where \(\chi_{\{x_i\}}^{-1}(U_i)\in \mathcal{B}([0,1])\). Hence \(\gamma^{-1}(U)\in \mathcal{B}([0,1])\) and \(\gamma\) is Borel-measurable.
Lemma 3. Let \(E\) be a locally convex space and \(\gamma_n:[0,1]\to E\) a sequence of Lusin-measurable functions and let \(\varepsilon>0.\) Then there exists a compact subset \(H_\varepsilon\) of \([0,1]\) such that \(\lambda([0,1]\setminus H_\varepsilon)<\varepsilon.\) Moreover, for each \(n\in \mathbb{N}\), the function \(\gamma_n|_{H_\varepsilon}\) is continuous.
Proof. Given \(n\in\mathbb{N}\), we consider a compact subset \(K_{\varepsilon,n}\subseteq [0,1]\) such that \(\lambda([0,1]\setminus K_{\varepsilon,n})<{\varepsilon}/{2^n}\) and \(\gamma_n|_{K_{\varepsilon,n}}\) is continuous. Next we define the set \[H_\varepsilon = \bigcap_{n=1}^\infty K_{\varepsilon,n}.\] Thus \(H_\varepsilon\) is compact in \([0,1]\) and \[\begin{align} \lambda\left( [0,1]\setminus H_\varepsilon \right) = \lambda\left( \bigcup_{n=1}^\infty [0,1]\setminus K_{\varepsilon,n} \right) \leq \sum_{n=1}^\infty \lambda([0,1]\setminus K_{\varepsilon,n}) \leq \sum_{n=1}^\infty \frac{\varepsilon}{2^n} = \varepsilon. \end{align}\] Moreover, since \(H_\varepsilon\subseteq K_{\varepsilon,n}\), each \(\gamma_n|_{H_\varepsilon}\) is continuous. ◻
We note that, under the assumption that the sequence \((\gamma_n)_n\) is a uniformly Cauchy sequence on an appropriate compact set, we obtain the following.
Lemma 4. Let \(E\) be a sequentially complete Hausdorff locally convex space and \((\gamma_n)_n\) be a sequence of Lusin-measurable functions. Assume that for each \(\varepsilon>0\), there exists a compact subset \(M_\varepsilon\) of \([0,1]\) such that \(\lambda([0,1]\setminus M_\varepsilon)<\varepsilon\) and \((\gamma_n|_{M_\varepsilon})_n\) is a uniformly Cauchy sequence in \(M_\varepsilon.\) Then the pointwise limit of \((\gamma_n)_n\) exists a.e. Moreover, the function \[\gamma:[0,1]\to E,\quad t\mapsto \gamma(t)=\left\{\begin{array}{rl} \lim_{n\to \infty} \gamma_n(t), & \text{if the limit exists.} \\ 0, & \text{if not.} \end{array}\right.\] is Lusin-measurable.
Proof. Let \(k \in \mathbb{N}\). By Lemma 3, there exists a compact subset \(H_k\) of \([0,1]\) such that \(\lambda([0,1]\setminus H_k )<1/2k\) and each \(\gamma_n|_{H_k}\) is continuous. By hypothesis, there exists a compact subset \(M_k\) of \([0,1]\) such that \(\lambda([0,1]\setminus M_k )<1/2k\) and \((\gamma_n)\) is uniformly Cauchy in \(M_k\), i.e., for each continuous seminorm \(q\) on \(E\), there exists \(n_{k,q}\in \mathbb{N}\) such that \[\sup_{t\in M_k} q\Big(\gamma_{m}(t)-\gamma_n(t)\Big)< \frac{1}{2k},\quad \forall m,n\geq n_{k,q}.\] We define the compact set \(F_k=H_k\cap M_k\), then \(\lambda([0,1]\setminus F_j)<1/k\) and \((\gamma_n|_{F_k})_n\) is a uniformly Cauchy sequence of continuous functions. Let us consider the Hausdorff locally convex space of continuous functions \(C(F_k,E)\) with the uniform convergence topology with respect to the continuous seminorms of \(E\). Since \(E\) is sequentially complete, the space \(C(F_k,E)\) is also sequentially complete. Then, the sequence \((\gamma_n|_{F_k})_n\) converges uniformly to a function \(\alpha_k \in C(F_k,E)\). We observe that if \(t\in F_{k_1}\cap F_{k_1}\), then \(\alpha_{k_1}(t)=\alpha_{k_2}(t)\) since they are the limit of \((\gamma_n(t))_n\) in the Hausdorff space \(E\). We define the measurable set \[F=\bigcup_{k=1}^\infty F_k.\] Then \[\lambda([0,1]\setminus F) \leq \lambda([0,1]\setminus F_k) \leq \frac{1}{k},\quad \forall k\in \mathbb{N}.\] Therefore \(\lambda([0,1]\setminus F)=0\). We define the function \[\gamma:[0,1]\to E,\quad t\mapsto \gamma(t)=\left\{ \begin{array}{rl} \alpha_k(t), & \text{for any k\in \mathbb{N} such that t\in F_k.} \\ 0, & \text{if t\not\in F.} \end{array}\right.\] Then the function \(\gamma:[0,1]\to E\) is well defined, it is the pointwise limit a.e. of the sequence \((\gamma_n)_n\) and is Lusin-measurable. Indeed, for each \(\varepsilon>0\), there exists a \(n_\varepsilon\in \mathbb{N}\) with \(1/n_\varepsilon\leq \varepsilon\) such that \(\lambda([0,1]\setminus F_{n_\varepsilon})<\varepsilon\) and \(\gamma|_{F_{n_\varepsilon}}=\alpha_{n_\varepsilon}\in C(F_{n_\varepsilon},E)\). ◻
Example 1. Let \(E\) be a sequentially complete Hausdorff locally convex space and \((\gamma_n)_n\) be a Cauchy sequence in \(C([0,1],E)\). If \(\varepsilon>0\), then applying Lemma 4 and assuming \(H_\varepsilon = [0,1]\), we have that the pointwise limit \(\gamma\) is Lusin-measurable.
Example 2. Consider the example of Remark 1. Then this sequence \((\gamma_n)_n\) does not verify the conditions of the Lemma 4. Indeed, fix \(x\in (0,1)\) and \(N_x\in \mathbb{N}\) be such that \(x+\frac{1}{2n}\in (0,1)\), for all \(n\geq N_x\). Then \[\begin{align} \gamma_n\left(x+\frac{1}{2n}\right)(x) = \max\left\{ 1-n\left\lvert x-\left(x+\frac{1}{2n}\right)\right\lvert, 0\right\}= \frac{1}{2}. \end{align}\] If \(m\geq 2n\geq N_x\), then \[\begin{align} \gamma_m\left(x+\frac{1}{2n}\right)(x) &= \max \left\{1-\frac{m}{2n}, 0 \right\} = 0. \end{align}\] Hence, if by considering the seminorm \(q_x(f)=\lvert f(x)\lvert\), we have that \[q_x \left(\gamma_m\left(x+\frac{1}{2n}\right) - \gamma_n\left(x+\frac{1}{2n}\right) \right) = \left\lvert 0-\frac{1}{2} \right\lvert=\frac{1}{2}.\] Let \(\varepsilon>0\), using the fact that each \(\gamma_n\) is continuous, we may assume that \(H_\varepsilon=[0,1]\). Then we have \[\frac{1}{2}\leq \sup_{t\in[0,1]} q_x \left(\gamma_m(t)-\gamma_n(t)\right),\quad \forall m\geq 2n\geq N_x.\]
Theorem 2. Let \(E\) be a sequentially complete Hausdorff locally convex space. If a function \(\gamma:[0,1]\to E\) is uniform limit a.e. of simple functions, then it is Lusin-measurable.
Proof. Let \((\beta_n)_n\) be a sequence of simple functions that converges uniformly a.e. to \(\gamma\), i.e., there exists a measurable set \(B\in\mathcal{B}([0,1])\) of measure zero, such that for each \(\varepsilon>0\) and each continuous seminorm \(q\) of \(E\), there exists \(n_{\varepsilon,q}\in \mathbb{N}\) such that \[\sup_{t\in [0,1]\setminus B} q(\gamma(t)-\beta_n(t))<\varepsilon/2\quad \forall n\geq n_{\varepsilon,q}.\] Denote by \(A=[0,1]\setminus B\), then we have that \[\sup_{t\in A} q(\beta_m(t)-\beta_n(t))<\varepsilon\quad \forall m,n\geq n_{\varepsilon,q}.\] Then by regularity of the Lebesgue measure, there exists a compact set \(K_\varepsilon\) of \([0,1]\) with \(K_\varepsilon\subseteq A\) such that \(\lambda(A\setminus K_\varepsilon)<\varepsilon/2\) and verifies \[\lambda([0,1]\setminus K_\varepsilon)\leq \lambda([0,1]\setminus A)+\lambda(A\setminus K_\varepsilon) < \varepsilon/2.\] Since each simple function is Lusin-measurable, by Lemma 3, there exists a compact subset \(H_\varepsilon\) of \([0,1]\) such that \(\lambda([0,1]\setminus H_\varepsilon)<\varepsilon/2\) and each \(\beta_n|_{H_\varepsilon}\) is continuous. Let us define \(F_\varepsilon=K_\varepsilon \cap H_\varepsilon\), then we have \[\begin{align} \lambda([0,1]\setminus F_\varepsilon) &\leq \lambda([0,1]\setminus K_\varepsilon) +\lambda([0,1]\setminus H_\varepsilon) < \varepsilon. \end{align}\] Since \(F_\varepsilon\subseteq K_\varepsilon\), we have \[\sup_{t\in F_\varepsilon} q(\beta_m(t)-\beta_n(t))<\varepsilon\quad \forall m, n\geq n_{\varepsilon,q}.\] and since \(F_\varepsilon\subseteq H_\varepsilon\), each \(\beta_n|_{F_\varepsilon}\) is continuous. Therefore, by Lemma 4, \(\gamma\) is Lusin-measurable since it coincides with the constructed pointwise limit. ◻
Example 3. For an open subset \(\Omega\) of \(\mathbb{R}^n\), the space of test functions on \(\Omega\), denoted by \[\mathcal{D}(\Omega)=\{ f\in C^\infty(\Omega) : \text{ \text{supp}(f) is compact in \Omega}\},\] is a sequentially complete Hausdorff locally convex space which is not metrizable.
Theorem 3. Let \(E\) be a sequentially complete Hausdorff locally convex space. Assume that \(E\) is metrizable and separable. Then the strongly-measurable function \(\gamma:[0,1]\to E\) is also Lusin-measurable.
Proof. Let \((\beta_n)\) be a sequence of simple functions that converges pointwise a.e. to \(\gamma\). Let \(\varepsilon>0\). Then by Lemma 3, there exists a compact subset \(K_\varepsilon\) of \([0,1]\) such that \(\lambda([0,1]\setminus K_\varepsilon)<\varepsilon/2\) and each \(\beta_n|_{K_\varepsilon}\) is continuous. Now considering \(E\) as metric space with metric \(d\) defined by the countable family of seminorms \(S=\{q_n : n\in \mathbb{N}\}.\) Then by Egorov’s Theorem, we have that for each \(\varepsilon>0\), there exists \(A_\varepsilon\in \mathcal{B}([0,1])\) with \(\lambda([0,1]\setminus A_\varepsilon)<\varepsilon/4\) such that \((\beta_n)_n\) converges uniformly to \(\gamma\) on \(A_\varepsilon\). Since the convergence of \((\beta_n)_n\) is uniform with respect to the metric \(d\), this fact implies that for each continuous seminorm \(q\in S\), there exists \(n_{\varepsilon,q}\in \mathbb{N}\) such that \[\sup_{t\in A_\varepsilon} q(\gamma(t)-\beta_n(t))<\varepsilon/2,\quad \forall n\geq n_{\varepsilon,q}.\] By regularity of the Lebesgue measure, there exists a compact subset \(H_\varepsilon\) of \([0,1]\) with \(H_\varepsilon\subseteq A_\varepsilon\) such that \(\lambda(A_\varepsilon\setminus H_\varepsilon)<\varepsilon/4\). Then \[\lambda([0,1]\setminus H_\varepsilon) \leq \lambda([0,1]\setminus A_\varepsilon) + \lambda(A_\varepsilon\setminus H_\varepsilon) < \varepsilon/2.\] Let \(F_\varepsilon=K_\varepsilon\cap H_\varepsilon\), then \[\lambda([0,1]\setminus F_\varepsilon)\leq \lambda([0,1]\setminus K_\varepsilon)+\lambda([0,1]\setminus H_\varepsilon) <\varepsilon,\] and each \(\beta_n|_{F_\varepsilon}\) is continuous. Moreover, since \(q(\beta_m(t)-\beta_n(t))\leq q(\beta_m(t)-\gamma(t))+q(\gamma(t)-\beta_n(t))\) and \(F_\varepsilon\subseteq H_\varepsilon\), we have \[\sup_{t\in F_\varepsilon} q(\beta_m(t)-\beta_n(t))<\varepsilon,\quad \forall m,n\geq n_{\varepsilon,q}.\] Then, by Lemma 4, the function \(\gamma\) is Lusin-measurable. ◻
Example 4. The space of smooth functions \(C^\infty(\mathbb{R}^n)\) is a separable Fréchet space.
Definition 2. Let \(E\) be a Hausdorff locally convex space and \(1\leq p<\infty\). We define the space \(\mathcal{L}^p([0,1],E)\) to be the set of all Lusin-measurable functions \(\gamma : [0,1]\to E\) such that for each continuous seminorm \(q\) on \(E\) we have that \[q\circ \gamma \in \mathcal{L}^p([0,1],\mathbb{R}).\]
Remark 4. We recall that a Lusin-measurable function \(\gamma:[0,1]\to E\) verifies \(\gamma(t)=0\) a.e. if, and only if, \(q(\gamma(t))=0\) a.e., for each continuous seminorm \(q\) on \(E\) (see e.g. [5], [10]).
Definition 3. Let \(E\) be a Hausdorff locally convex space. If \(\gamma,\eta:[0,1]\to E\) are two Lusin-measurable functions, we say \(\gamma \sim \eta\) if and only if \(\gamma(t) = \eta(t)\) for almost all \(t\in [0,1]\) and write \([\gamma]\) for the
equivalence class of \(\gamma\). We define the real vector space \[L^p([0,1],E)= \mathcal{L}^p([0,1],E) / [0].\] For each continuous seminorm \(q\) of \(E\), we define the seminorm \[\lVert [\gamma]\lVert_{\mathcal{L}^p,q}=\left( \int_0^1 q(\gamma(t))^pdt \right)^{1/p},\quad \forall [\gamma]\in L^p([0,1],E).\] Then these seminorms define a
Hausdorff locally convex topology for \(L^p([0,1],E)\).
If no confusion arises, we will denote \([\gamma]\) simply as \(\gamma\). For the case \(p=\infty\), using the essential supremum, it is possible to define
the space \(L^\infty([0,1],E)\), but we will only focus on the finite case.
Remark 5. We recall the topological embeddings \[C([0,1],E)\hookrightarrow L^\infty([0,1],E) \hookrightarrow L^r([0,1],E)\hookrightarrow L^p([0,1],E)\hookrightarrow L^1([0,1],E),\] where \(1\leq p\leq r <\infty\) (see e.g. [5]). Moreover, in [10], the authors show that if \(E\) is sequentially complete, the space \(L^\infty([0,1],E)\) is sequentially complete. Furthermore, if \(E\) is quasi-complete and fundamentally \(L^p\)-bounded, then \(L^p([0,1],E)\) is locally complete for \(1\leq p<\infty\).
Theorem 6. Let \(1\leq p<\infty\) and let \(E\) be a Hausdorff locally convex space.
If \(\gamma\in L^p([0,1],E)\), then for each \(\varepsilon>0\) and each continuous seminorm \(q\), there exists a simple function \(\beta_{\varepsilon,q}:[0,1]\to E\) such that \[\lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q} < \varepsilon.\] Consequently, the space of simple functions \(\mathcal{S}([0,1],E)\) is dense in \(L^p([0,1],E)\).
Proof. Let \(q\) be a continuous seminorm of \(E\) and \(\varepsilon>0\). By absolute continuity of the integral, there exists \(\delta=\delta_{\gamma,\varepsilon,q}>0\) such that for every measurable set \(A\in \mathcal{B}([0,1])\) \[\lambda(A)<\delta \implies \int_A q(\gamma(t))^pdt<
\frac{\varepsilon^p}{2}.\] Moreover, by Lemma 2, for \(\theta_\varepsilon=\min\left\{\delta,\varepsilon/\sqrt[p]{2}\right\}\) there exists a compact
subset \(K_{\varepsilon}\) of \([0,1]\) and a simple function \(\beta_{\varepsilon,q}:[0,1]\to E\) such that \(\lambda([0,1]\setminus K_\varepsilon)<\theta_\varepsilon\), the function \(\gamma|_{K_\varepsilon}\) is continuous and \[\sup_{t\in K_\varepsilon}
q(\gamma(t)-\beta_{\varepsilon,q}(t))<\theta_\varepsilon \leq\frac{\varepsilon}{\sqrt[p]{2}}\] Without loss of generality, we suppose that \(\beta_{\varepsilon,q}(t)=0\) for all \(t\in
[0,1]\setminus K_\varepsilon\). Then \[\begin{align} \lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q}^p &= \int_0^1 q(\gamma(t)-\beta_{\varepsilon,q}(t))^p dt \\ & =
\int_{K_\varepsilon}q(\gamma(t)-\beta_{\varepsilon,q}(t))^p dt + \int_{[0,1]\setminus K_\varepsilon} q(\gamma(t)-\beta_{\varepsilon,q}(t))^pdt\\ &\leq \sup_{t\in K_\varepsilon} q(\gamma(t)-\beta_{\varepsilon,q}(t))^p\lambda(K_{\varepsilon}) +
\int_{[0,1]\setminus K_\varepsilon} q(\gamma(t))^p dt\\ & \leq \frac{\varepsilon^p}{2}+\frac{\varepsilon^p}{2} = \varepsilon^p.
\end{align}\] Hence \(\lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q} < \varepsilon\). Let \(\gamma\in L^p([0,1],E).\) Then a basic open \(\gamma\)-neighborhood is a set of the form \[V_\gamma = \gamma+\bigcap_{i=1}^n \lVert \cdot \lVert_{\mathcal{L}^p,q_i}^{-1}\left([0,\varepsilon)\right)\] for the continuous seminorms \(\varepsilon>0\) and \(q_1,...,q_n\) of \(E\).
Next, we define the continuous seminorm \(q:E\to [0,\infty)\) by \[q(x)=\max\{q_1(x),...,q_n(x)\},\quad \forall x\in E\] Then, there exists \(\beta_{\varepsilon,q}\in \mathcal{S}([0,1],E)\) such that \[\lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q}<\varepsilon.\] Therefore, for each \(i\in
\{1,...,n\}\), we have \[\lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q_i} =\int_0^1 q_i (\gamma(t)-\beta_{\varepsilon,q}(t))dt \leq\int_0^1 q(\gamma(t)-\beta_{\varepsilon,q}(t))dt = \lVert
\gamma-\beta_{\varepsilon,q} \lVert_{\mathcal{L}^p,q}< \varepsilon.\] Therefore \(\beta_{\varepsilon,q}\in V_\gamma\) and \(\mathcal{S}([0,1],E)\) is dense. ◻
We recall [13].
Lemma 5 (Urysohn’s Lemma). Let \(X\) be a locally compact Hausdorff space. Suppose that \(V\subseteq X\) is open and \(K\subseteq V\) compact. Then there exists a continuous function with compact support \(f:X\to \mathbb{R}\) such that \[\chi_K \leq f \leq \chi_V.\]
Theorem 7. Let \(1\leq p<\infty\) and \(E\) be a Hausdorff locally convex space. Then the space of continuous functions \(C([0,1],E)\) is dense in \(L^p([0,1],E)\).
Proof. Let \(y_0\in E\) and \(A\in\mathcal{B}([0,1])\). Since \(\mathcal{S}([0,1],E)\) is dense in \(L^p([0,1],E)\). Then it suffices to show that the simple function \(y_0\chi_A:[0,1]\to E\) belongs to the closure of \(C([0,1],E)\).
In fact, \[\begin{align} \lambda(A)&=\sup \{\lambda(K) : \text{K is compact and K\subseteq A}\}\\ &= \inf \{ \lambda(U) : \text{U is open and A\subseteq U}\}.
\end{align}\] Then there exists a sequence of compact sets \((K_n)_n\) and open sets \((U_n)_n\) such that for each \(n\in \mathbb{N}\), we have that
\(K_n\subseteq A\subseteq U_n\) and \[\lambda(A\setminus K_n) <\frac{1}{2n}\quad \land \quad \lambda(U_n\setminus A)<\frac{1}{2n}.\] By Urysohn’s Lemma, there exists a continuous
function \(f_n:[0,1]\to \mathbb{R}\) such that \[\chi_{K_n}\leq f_n\leq \chi_{U_n}.\] Then the function \[\alpha_n:[0,1]\to E,\quad \alpha_n(t)=f_n(t)
y_0,\] is in \(C([0,1],E)\).
We see that if \(t\in K_n\) or \(t\in [0,1]\setminus U_n\) then \((\alpha_n - y_0\chi_{A})(t)=0\). Now, since \[U_n\setminus K_n =
(U_n\setminus A)\cup (A\setminus K_n).\] Therefore, for each continuous seminorm \(q\) of \(E\) we have that \[\begin{align} \lVert \alpha_n-y_0\chi_A
\lVert_{\mathcal{L}^p,q}^p &= \int_0^1 q( \alpha_n(t)-y_0\chi_A(t))^pdt \\ &= \int_{U_n\setminus K_n} q(f_n(t) y_0 - y_0 \chi_A(t))^p dt \\ & = \int_{U_n\setminus A} q(f_n(t) y_0 - y_0 \chi_A(t))^p dt + \int_{A\setminus K_n} q(f_n(t) y_0 - y_0
\chi_A(t))^p dt \\ &= q( y_0 )^p\left( \int_{U_n\setminus A} \lvert f_n(t)\lvert^p dt + \int_{A\setminus K_n} \lvert f_n(t) -1\lvert^p dt\right) \\ &\leq q(y_0)^p\left( \lambda(U_n\setminus A) + \lambda(A\setminus K_n) \right) \\ &\leq q(y_0)^p
\left(\frac{1}{2n}+ \frac{1}{2n}\right) = \frac{q(y_0)^p}{n}.
\end{align}\] Hence \(\alpha_n \to y_0\chi_A\) as \(n\to \infty\) in \(L^p([0,1],E)\). ◻
Remark 8. Let us consider the locally convex space \(E=\{f:\mathbb{R}\to \mathbb{R}\}\) and the sequence \((\gamma_n)_n\) given in Remark 1. Then \((\gamma_n)_n\) is in \(L^p([0,1],E)\). We claim that this sequence converge in the \(L^p([0,1],E)\) norm. In fact, let be \(0_e:\mathbb{R}\to \mathbb{R}\) and \(0_E:[0,1]\to E\) be the respective constants maps \(0.\) It is clearly that \(0_E\in L^p([0,1],E).\) Moreover, for \(x_0\in \mathbb{R}\) fixed we have that, \[\begin{align} \lVert \gamma_n - 0_E\lVert_{\mathcal{L}^p,q_{x_0}}^p &= \int_0^1 (q_{x_0} ( \gamma_n (t) ))^p dt \\ &= \int_0^1 (\gamma_n (t)(x_0))^p dt\\ &= \int_0^1 \max \{ 1-n\lvert x_0-t\lvert, 0\}^p dt\\ &= \int_{\left[x_0-\frac{1}{n},x_0+\frac{1}{n}\right]\cap [0,1]} \left( 1-n\lvert x_0 -t \lvert\right)^p dt \\ &\leq \int_{\left[x_0-\frac{1}{n},x_0+\frac{1}{n}\right]\cap [0,1]} 1 dt \\ &\leq \frac{2}{n}. \end{align}\] Hence \(\gamma_n \to 0_E\) in \(L^p([0,1],E)\). On the other hand, let \(\gamma:[0,1]\to E\) be the pointwise limit of the sequence \((\gamma_n)_n\) (see Remark 1). Then, we notice that \[\lambda\Big(\{t\in [0,1] : \gamma(t)\neq0_E(t)\}\Big)=1.\] Thus, the pointwise limit is not equal a.e. to the limit in \(L^p([0,1],E)\). Hence for functions with values in a non-metrizable locally convex space, the convergence in \(L^p\) does not imply the existence of a subsequence which converges to the pointwise limit. This shows that functions taking values in a non-metrizable space may exhibit behaviors that do not occur in the metrizable setting.
Definition 4. Let \(E\) be a Hausdorff locally convex space and \(E'\) its topological dual space. We say that a function \(\gamma:[0,1]\to E\) is weak-integrable if there exists \(z\in E\) such that for each \(f\in E'\), we have \(f\circ \gamma\in \mathcal{L}^1([0,1],\mathbb{R})\) and \[f(z) = \int_0^1 f(\gamma(t))dt,\quad \forall f\in E'.\] This element \(z\) is necessarily unique and we denote \(z=\int_0^1 \gamma(t)dt\).
The following result ([5]) shows a sufficient condition for the existence of a weak-integral.
Theorem 9. Let \(E\) be a sequentially complete Hausdorff locally convex space. Then each \(\gamma\in \mathcal{L}^1([0,1],E)\) is weak-integrable. Furthermore, the function \[\eta:[0,1]\to E,\quad \eta(t)=\int_0^t \gamma(s)ds\] is continuous.
We recall [8].
Lemma 6. For each \(n\in \mathbb{N}\), we denote the dyadic partition \(\pi_n\) of \([0,1],\) that is, \[\pi_n= \left\{ \left[\frac{k-1}{2^n},\frac{k}{2^n}\right] : k\in\{1,....,2^n\}\right\}.\] Let \(A\in \mathcal{B}([0,1])\). Then for each \(\varepsilon>0\), there exists an \(n_\varepsilon\in \mathbb{N}\) and a finite collection \(I_1,...,I_k\in \pi_{n_\varepsilon}\), such that \[\lambda\left(A\triangle\bigcup_{j=1}^k I_j\right)<\varepsilon,\] where \(\triangle\) denotes the symmetric difference.
Remark 10. Let \(\gamma\in L^p([0,1],E)\). By the previous lemma and the absolute continuity of the integral, for each \(\varepsilon>0\) and each \(A\in \mathcal{B}([0,1])\), there exists \(n_{\varepsilon,A}\in \mathbb{N}\) for which there exists a finite collection \(I_1,...,I_k\in \pi_{n_{\varepsilon,A}}\) satisfying \[\int_{A\triangle\bigcup_{j=1}^k I_j} q(\gamma(t))^p dt < \varepsilon.\]
We recall [4].
Lemma 7. Let \(E\) be a locally convex space and \(\gamma:[0,1]\to E\) be a weak-integrable function. Then, for each continuous seminorm \(q\) of \(E\) we have that \[q\left( \int_0^1 \gamma(t)dt \right) \leq \int_0^1 q(\gamma(t))dt.\]
A function \(\gamma:[0,1]\to E\) is \(\lambda\)-measurable if for each \(\varepsilon>0\), there exists \(A_\varepsilon\in \mathcal{B}([0,1])\) such that \(\lambda([0,1]\setminus A_\varepsilon)<\varepsilon\) and \(\gamma|_{A_\varepsilon}\) is uniform limit of simple functions. In [11], the authors state a density result using the notion of \(\lambda\)-measurability with values in a quasi-complete Hausdorff locally convex space. In our next result we considered the hypothesis of sequentially completeness instead of quasi-completeness used by [11].
Theorem 11. Let \(E\) be a sequentially complete Hausdorff locally convex space, \(1\leq p <\infty\) and \(\gamma\in L^p([0,1],E)\). For \(n\in \mathbb{N}\), let \(\pi_n\) the dyadic partition of \([0,1]\) and we define the sequence of simple functions by \[\gamma_n:[0,1]\to E,\quad \gamma_n(t) = \sum_{j=1}^{2^n} \left( 2^n \int_{I_j} \gamma(t)dt\right) \chi_{I_j}(t).\] Then the sequence \((\gamma_n)_n\) converges to \(\gamma\) in \(L^p([0,1],E)\).
Proof. Under the hypothesis that \(E\) is sequentially complete. Then by Theorem 9, the simple function \(\gamma_n\) are well defined. Since \(\int_{I_j}\gamma(t)dt\) exists. Moreover, they are Lusin-measurable and \(L^p\)-integrable.
Now, let \(\varepsilon>0\) and let \(q\) be a continuous seminorm of \(E\). Then there exists a simple function \(\beta_{\varepsilon,q}:[0,1]\to E\) such that \(\lVert \gamma-\beta_{\varepsilon,q}\lVert_{\mathcal{L}^p,q} < \varepsilon/3\). This simple function \(\beta_{\varepsilon,q}\) is weak-integrable and if \[\beta_{\varepsilon,q}=\sum_{i=1}^m y_i \chi_{A_i},\] where \(y_1,...,y_m\in E\) and \(A_1,...,A_m\in \mathcal{B}([0,1])\) form a measurable partition of \([0,1]\), then \[\int_0^1 \beta_{\varepsilon,q}(t)dt = \sum_{i=1}^m y_i \lambda(A_i).\] Thus
we have that \[\begin{align} \lVert \gamma-\gamma_n \lVert_{\mathcal{L}^p,q} &\leq \lVert \gamma-\beta_{\varepsilon,q} \lVert_{\mathcal{L}^p,q} + \lVert \beta_{\varepsilon,q} - \gamma_n \lVert_{\mathcal{L}^p,q}.
\end{align}\] The first term on right hand side of the above inequality is already bounded by \(\varepsilon/3.\) Thus, we only need to estimate the second term, \[\begin{align} \lVert
\beta_{\varepsilon,q} - \gamma_n \lVert_{\mathcal{L}^p,q}^p &= \int_0^1 q\left( \sum_{i=1}^m y_i \chi_{A_i}(s)- \sum_{j=1}^{2^n} \left( 2^n \int_{I_j} \gamma(t)dt\right) \chi_{I_j}(s) \right)^p ds \\ &= \sum_{i=1}^m \int_{A_i} q\left(
y_i\chi_{A_i}(s)- \sum_{j=1}^{2^n} \left( 2^n \int_{I_j} \gamma(t)dt\right) \chi_{I_j}(s) \right)^p ds.
\end{align}\] By Remark 10, let \(n_{\varepsilon,A_1,...,A_m}\in \mathbb{N}\) be such that for each \(i\in
\{1,....,m\}\), there exists a finite collection of intervals \(I_{i,1},...,I_{i,k_i}\in \pi_{n_{\varepsilon,A_1,...,A_m}}\) that verify \[\sum_{i=1}^m \int_{A_i \triangle \cup_{j=1}^{k_i}
I_{i,j}} q\left( y_i\chi_{A_i}(s)- \sum_{j=1}^{2^n} \left( 2^n \int_{I_j} \gamma(t)dt\right) \chi_{I_j}(s) \right)^p ds < \varepsilon/3\] and \[\sum_{i=1}^m q(y_i)\lambda\Big(A_i \triangle \cup_{j=1}^{k_i} I_{i,j}\Big)
\leq \varepsilon/3.\] Since \[A_i \subseteq \left( A_i\cap \bigcup_{j=1}^{k_i} I_{i,j}\right) \cup \left( A_i \triangle \bigcup_{j=1}^{k_i} I_{i,j}\right).\] Hence for \(n\geq
n_{\varepsilon,A_1,...,A_m}\) we obtain that, \[\begin{align} \lVert \beta_{\varepsilon,q} - \gamma_n \lVert_{\mathcal{L}^p,q}^p &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} \int_{A_i\cap I_{i,j}} q\left( y_i\chi_{A_i}(s)-
\left( 2^n \int_{I_{i,j}} \gamma(t)dt\right) \chi_{I_j}(s) \right)^p ds + \varepsilon/3 \\ &= \sum_{i=1}^m \sum_{j=1}^{k_i} q\left( y_i- 2^n \int_{I_{i,j}} \gamma(t)dt \right)^p \lambda\left(A_i\cap I_{i,j}\right) + \varepsilon/3 \\ &\leq
\sum_{i=1}^m \sum_{j=1}^{k_i} q\left( y_i- 2^n \int_{I_{i,j}} \gamma(t)dt \right)^p \frac{1}{2^n} + \varepsilon/3 \\ &= \sum_{i=1}^m \sum_{j=1}^{k_i} q\left( 2^n \int_{ I_{i,j}} y_i\chi_{I_{i,k}}(t) dt- 2^n \int_{I_{i,j}} \gamma(t)dt \right)^p
\frac{1}{2^n} + \varepsilon/3 \\ &= \sum_{i=1}^m \sum_{j=1}^{k_i} q\left( 2^n \int_{ I_{i,j}} y_i\Big(\chi_{A_i\cap I_{i,k}}(t) + \chi_{I_{i,k}\setminus A_i}(t)\Big)dt- 2^n \int_{I_{i,j}} \gamma(t)dt \right)^p \frac{1}{2^n} + \varepsilon/3 \\ &=
\sum_{i=1}^m \sum_{j=1}^{k_i} q\left( 2^n \left(\int_{ I_{i,j}} (y_i\chi_{A_i}(t)dt -\gamma(t))dt\right) + 2^n \int_{ I_{i,j}} y_i\chi_{I_{i,k}\setminus A_i}(t)dt\right)^p \frac{1}{2^n} + \varepsilon/3 \\ &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} q\left(
\int_{ I_{i,j}} (y_i\chi_{A_i}(t)dt -\gamma(t))dt\right)^p \frac{2^{pn}}{2^n} + q\left(\int_{ I_{i,j}} y_i\chi_{I_{i,k}\setminus A_i}(t)dt\right)^p \frac{2^{pn}}{2^n} + \varepsilon/3 \\ &= \sum_{i=1}^m \sum_{j=1}^{k_i} q\left( \int_{ I_{i,j}}
(\beta_{\varepsilon,q}(t)dt -\gamma(t))dt\right)^p \frac{2^{pn}}{2^n} + q\left(\int_{ I_{i,j}} y_i\chi_{I_{i,k}\setminus A_i}(t)dt\right)^p \frac{2^{pn}}{2^n} + \varepsilon/3.
\end{align}\] Under the condition \(\frac{1}{p}+\frac{1}{r}=1.\) Then by Lemma 7 we have \[\begin{align}
\sum_{i=1}^m \sum_{j=1}^{k_i} q\left( \int_{I_{i,j}} \left(\beta_{\varepsilon,q}(t) - \gamma(t) \right)dt \right)^p \frac{2^{pn}}{2^n} &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} \left( \int_{I_{i,j}} q\left( \beta_{\varepsilon,q}(t)-\gamma(t)\right) dt
\right)^p \frac{2^{pn}}{2^n}\\ &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} \left( \int_{I_{i,j}} q(\beta_{\varepsilon,q}(s)-\gamma(s))^p dt \right)^{p/p} \left(\int_{I_{i,j}} 1 dt \right)^{p/r} \frac{2^{pn}}{2^n}\\ &= \sum_{i=1}^m \sum_{j=1}^{k_i} \left(
\int_{I_{i,j}} q(\beta_{\varepsilon,q}(s)-\gamma(s))^p dt \right)\frac{2^{pn}}{2^{np/r}2^n}\\ &= \left(\int_0^1 q(\beta_{\varepsilon,q}(s)-\gamma(s))^p dt\right) 2^0\\ & \leq \varepsilon/3.
\end{align}\] On the other hand \[\begin{align} \sum_{i=1}^m \sum_{j=1}^{k_i} q\left(\int_{ I_{i,j}} y_i\chi_{I_{i,k}\setminus A_i}(t)dt\right)^p \frac{2^{pn}}{2^n} &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} \left( \int_{
I_{i,j}} q\Big(y_i\chi_{I_{i,k}\setminus A_i}(t)\Big)dt\right)^p \frac{2^{pn}}{2^n} \\ &\leq \sum_{i=1}^m \sum_{j=1}^{k_i} \left( \int_{ I_{i,j}} q\Big(y_i\chi_{I_{i,k}\setminus A_i}(t)\Big)dt \right)^{p/p} \left(\int_{I_{i,j}} 1 dt \right)^{p/r}
\frac{2^{pn}}{2^n}\\ \quad \quad &\leq \sum_{i=1}^m \int_{\cup_{j=1}^{k_i}I_{i,j}} q\Big(y_i\chi_{\cup_{j=1}^{k_i}I_{i,k}\setminus A_i}(t)\Big)dt \\ & = \sum_{i=1}^m \int_{\cup_{j=1}^{k_i}I_{i,k}\setminus A_i}
q(y_i)\chi_{\cup_{j=1}^{k_i}I_{i,k}\setminus A_i}(t)dt \\ &\leq \sum_{i=1}^m q(y_i)\lambda\Big(A_i \triangle \cup_{j=1}^{k_i} I_{i,j}\Big)\\ &\leq \varepsilon/3.
\end{align}\] Therefore \[\lVert \gamma -\gamma_n \lVert_{\mathcal{L}^p,q} \leq \varepsilon,\quad \forall n\geq n_\varepsilon.\] ◻
Supported by DICYT-USACH (grant 042632PC-POSTDOC)↩︎