I need help proving that the set $B \cap L^2\Big( (0,T) \times (0,1)\Big)$ is a closed subset of $L^2\Big( (0,T) \times (0,1)\Big)$, where $B$ is defined as:
$$B=\Big\{x \in L^{\infty}(0,T;L^1(0,1)) \ : \|x(t)\|_{L^1(0,1)} \le 1 , \ \ \text{a.a in } \ (0,T) \Big\} $$
Any suggestions or proofs would be greatly appreciated!
Let $(x_n(\cdot,\cdot))_n$ be a sequence in $B \cap L^2\Big((0, T) \times (0,1)\Big)$ which converges in $L^2\Big((0, T) \times (0,1)\Big)$ to some function $x(\cdot,\cdot)$. All we have to show is that, for almost all $t \in (0,T)$: $\|x(t, \cdot)\|_{L^1(0,1)} \leq 1$.
Since $(x_n)_n$ converges in $L^2\Big((0, T) \times (0,1)\Big)$ to $x$, the sequence of functions $\left(t \mapsto x_n(t, \cdot)\right)_n$ converges in $L^2\Big((0,T); L^1(0,1)\Big)$ to $t \mapsto x(t, \cdot)$ (left as a quick exercise).
Therefore, there exists a subsequence $\left(t \mapsto x_{\varphi(n)}(t, \cdot)\right)_n$ converging pointwise to $t \mapsto x(t,\cdot)$ outside of a set $N_{-1}$ of measure $0$. Moreover, let $N_n$, for $n \in \mathbb{N}$, be a set of measure $0$ for which: $$\forall t \in (0, T) \setminus N_n,\quad \|x_{\varphi(n)}(t, \cdot)\|_{L^1(0,1)} \leq 1$$ We now consider this, where $N := \bigcup_{n = -1}^{+\infty} N_n$ is still of measure $0$ as a countable union of $0$ measure sets: $$\forall t \in (0,T) \setminus N,\quad \|x(t, \cdot)\|_{L^1(0,1)} \leq \|x(t, \cdot) - x_{\varphi(n)}(t, \cdot)\|_{L^1(0,1)} + \|x_{\varphi(n)}(t,\cdot)\|_{L^1(0,1)}$$ Since $\|x(t, \cdot)\|_{L^1(0,1)}$ is constant in regards to $n$, we can take the $\limsup_{n \to \infty}$ in the RHS, which provides, for $t \notin N$: $$\begin{split}\|x(t, \cdot)\|_{L^1(0,1)} &\leq \limsup_{n \to \infty}\left(\|x(t, \cdot) - x_{\varphi(n)}(t, \cdot)\|_{L^1(0,1)} + \|x_{\varphi(n)}(t,\cdot)\|_{L^1(0,1)}\right)\\ &\leq \limsup_{n \to \infty}\underbrace{\left(\|x(t, \cdot) - x_{\varphi(n)}(t, \cdot)\|_{L^1(0,1)}\right)}_{\xrightarrow[n \to \infty]{}\,\, 0 \,\text{since } t \,\notin\, N_{-1}} + \limsup_{n \to \infty}\underbrace{\left(\|x_{\varphi(n)}(t,\cdot)\|_{L^1(0,1)}\right)}_{\leq\,1 \,\text{since } t \,\notin\, N_{}}\\ &\leq 0 + 1 = 1\end{split}$$ This lets us conclude that $B \cap L^2\Big((0, T) \times (0,1)\Big)$ is indeed closed in $L^2\Big((0, T) \times (0,1)\Big)$.