Докажите включение - Tеория меры

[Solunac]

Если $S $является замкнутое подпространство в Банаховом пространстве $L^1 (\mu)$, где $\mu$ ee $\sigma$-конечная мера на пространстве $X.$ Пусть $f\in S$ подразумевает $f\in L^p (\mu)$ для некоторого $p>1$. Докажите, что существует $q>1$, такие что $S\subset L^q (\mu)$.

I don't speak Russian very well, but understand it (write suggestions on Russian). I hope you understand what is the problem (I tried). Спасибо.

[Юстас]

Think about Baire Category Theorem..

[Solunac]

Ошибка. Простите. $\mu (X)<\infty.$ $\mu$ это конечная мера на пространстве $X.$ Извините...

Другая вещь, которая может быть связана с проблемой. Мы знаем, что $L^p(\mu)\subset L^q(\mu)$ за $p > q\geqslant 1.$ При каких условиях эти два пространства содержат те же функции?

[Хорхе]

При условии, что $\mu$ сосредоточена на конечном числе атомов.

[Solunac]

Hi again. Here is what I have done.

Because $S=\bar{S}$ than $S$ is complete subspace of $L^1(\mu)$.

Let $S^p := \{f\in S : f\in L^p(\mu)\}$ for $p>1.$
From $\mu (X)<\infty$ we have that - $L^q(\mu)\subset L^p(\mu)$ for some $q, p$ set to $q>p$ and $q,p>1$. That implicates $S^q\subset S^p$ for $q>p$.
Now we can make sequence $\{S^p\}$ just to use that indexes $p$ which are in $\{2,3,4,\dots\}$ and in $\{1\frac{1}{2}, 1\frac{1}{3}, 1\frac{1}{4}, \dots\}$ (we don't need all $(1,\infty]$).
So let that be index set $I.$$|I|=|\mathbb{N}|$.
We now have that $S=\cup_{p\in I} S^p$. From this we can even form a disjoint family of subsets $D^p$ so that $S=\cup_{p\in I} D^p$.

By the one version of Baire Category Theorem, there exists $q\in I$ so that closure $\bar{S^q}$ has an interior point (we can say that for $D^q$ respectively). Let that interior point (function) be $g\in S^q$. That means that it has a ball around itself for some $\epsilon >0$ - $B(g,\epsilon)\subset \bar{S^q}$ and of course $B(g,\epsilon)\subset S^q$ too.

Now all elements in $S^q$ $(D^q$ if we are looking on it simultaneously) can be written like linear combination of some element from $B(g,\epsilon)$.

But now what?

We can assume opposite that exists $f\in S^p$ where $1<p<q$ and $f\notin S^q$ (same thing can be said for $D^p$ and $D^q$). But again, I don't see anything useful with I can do something... So, please help.

I don't speak Russian very well, but understand it (freely write suggestions on Russian).
Спасибо.