If $X\subset L^1$ is a closed vector space and $X\subset \bigcup_{11$. This is exercise 4.8 in Brezis's Functional Analysis book. Let $\Omega$ be a $\sigma$-finite measure space and $X\subset L^1(\Omega)$ be a closed vector space with the property
$$X\subset \bigcup_{1<p\leq\infty }L^p(\Omega).$$
We need to show that there is a $p>1$ such $X\subset L^p(\Omega)$. Here is the solution as given in the book:

Let  $X_n=\{f\in X\cap L^{1+1/n}(\Omega):\|f\|_{1+1/n}\leq n\}.$ Then
$X=\bigcup_{n=1}^\infty X_n$. Indeed, if $f\in X$ then $f\in L^q(\Omega)$ for some $q>1$ and hence by interpolation $f\in L^{1+1/n}(\Omega)$ for all $n$ such that $1+1/n\leq q$ and moreover,
$$\|f\|_{1+1/n}\leq\|f\|_1^{\alpha_n}\cdot\|f\|^{1-\alpha_n}_q, \;\;\text{ with }\;\;\frac{1}{1+1/n}=\frac{\alpha_n}{1}+\frac{1-\alpha_n}{q}.$$  By
Baire category theorem, there is an $X_{n_0}$ with non-empty interior.
This means that $X\subset L^{1+1/n_0}(\Omega).$

The last sentence I don't get. First to apply the Baire category, we needed to have shown that $X_n$ is closed in $X$, which is not there. Second, why does $\text{Int}(X_{n_0})\neq\emptyset$ imply that $X\subset L^{1+1/n_0}(\Omega)?$ Is it related to the fact that $X$ is vector space?
 A: Closedness of $X_n$'s follows immediately from Fatou's Lemma: If $\|f_j\|_{1+\frac 1 n} \leq n$ for all $j$ and $f_j \to f$ in $X$ then there is a  subsequence $f_{j_i}$ which converges a.e. so $\|f\|_{1+\frac 1 n} \leq  \lim \inf_k \|f_{j_k}\|_{1+\frac 1 n} \leq n$.
Suppose there is an open ball $B(f_0,r)$ in $X$ which is $ \subset X_{n_0}$. Let $f \in X$. Then $\|f_0+\frac  f N\|_{1+\frac 1 {n_0}} \leq n_0$ for $N$ sufficiently large (since $f_0+\frac  f N \in B(f_0,r)$). This implies that $f_0+\frac  f N\in L^{1+\frac 1 {n_0}}(\Omega)$ and we also have $f_0 \in L^{1+\frac 1 {n_0}}(\Omega)$. Since $L^{1+\frac 1 {n_0}}(\Omega)$ is a vector space it follows that $f \in L^{1+\frac 1 {n_0}}(\Omega)$.
A: *

*Being a closed linear subspace of $L_1$, $X$ is itself a Banach space.


*Each $X_n$ is a closed subset of $X$. Proof. Suppose $f\in \overline{X}_n^X$ (the closure of $X_n$ relative to $X$). Then there is a sequence $\{f_k:k\in\mathbb{N}\}\subset X_n$ such that $\|f-f_k\|_{L_1}\xrightarrow{k\rightarrow 0}0$. By standard results (application of Chebyshev Markov, and convergence in measure type if results (here is where $\sigma$-algebra is used) there is a subsequence $f_{k_m}$ that converges to $f$ $\mu$-a.s. An application of Fatou's lemma implies that $$\int|f|^{1+\tfrac{1}{n}}\,d\mu\leq \liminf_m\int|f_{k_m}|^{1+\tfrac1n}\,d\mu\leq n^{1+\tfrac1n}$$ Hence $f\in X_n$.


*As $X$ is a complete metric space  and $X$ is the countable union of closed subsets $X_n$ of $X$, then there is $n_0$ such that $\operatorname{int}(X_{n_0})\neq\emptyset$. (Baire's category theorem)


*If $B_X(f; r)+\{g\in X:\|f-g\|_{L_1}<r\}\subset X_{n_0}$, then $B(0;r)_X= B_X(f; r)-f$ is a an open ball in $X$ around $0$. The conclusion follows by noticing that in any Banach space $X$, $X=\bigcup_n nB_X(0;r)$ where $B_X(0;r)$ is an open ball around $0$ in $X$.
