In *Antoine Henrot Michel Pierre* - 
**Variation et optimisation de formes, Une analyse geometrique**, a book I'm studying I found an interesting problem. The problem is listed below. The first 3 points of the problem are pretty easy, and I solved them. The 4-th seems a little harder. The only indication I get is to use point 3) and the Baire theorem for $(\Sigma,\delta)$.



> 
> Denote by $\Sigma$ the quotient space
>  of the family of Lebesgue measurable
>  sets of $\Bbb{R}^N$ by the equivalence
>  relation $E\_1 \sim E\_2 \Leftrightarrow\chi\_{E\_1}=\chi\_{E\_2} a.e.$. Denote by
>  $|X|$ the Lebesgue measure of the
>  measurable set $X$.
> 
> 
> 1) Prove that
>  $\delta(E\_1,E\_2)=\arctan( |E\_1 \Delta E\_2|)$ is a distance on $\Sigma$.
> 
> 
> 2) Prove that given $(E\_n)\_{n \geq 1}, E$ measurable sets in $\Bbb{R}^N$ the
>  following three properties are
>  equivalent.
> 
> 
> * $\delta(E\_n,E) \to 0$;
> * $\chi\_{E\_n}-\chi\_E \xrightarrow{\sigma(L^1,L^\infty)} 0$;
> * $\chi\_{E\_n}-\chi\_E \xrightarrow{L^1} 0$.
> 
> 
> 3) Prove that $(\Sigma,\delta)$ is a
>  complete metric space.
> 
> 
> 4) Given the sequence $ (f\_n)$ of
>  integrable real valued functions on
>  $\Bbb{R}^N$, such that for any
>  measurable set $E$ of $\Bbb{R}^N$
>  there exists $\displaystyle \lim\_{n\to \infty}\int\_E f\_n$, prove that if
>  $|E| \to 0$ then $\displaystyle\sup\_n\int\_E |f\_n| \to 0$.
>  (Hint: Use the Baire category theorem for $(\Sigma,\delta)$)
> 
> 
> 


The question is: How can I apply Baire theorem to solve the 4-th point in the problem?



Baire Category Theorem Application
Choose $\varepsilon > 0$ and consider sets defined by
$$\Sigma\_k = \{E:\ \left|\int\limits\_{E} (f\_n-f\_m) \right| \leqslant \varepsilon, \textrm{ if } n,m \geqslant k \}$$
Since for any measurable set a limit of integrals exists, we have
$\Sigma = \bigcup\limits\_{k} \Sigma\_k$. Note, that given an integrable function $f$, the functional $f(E):= \int\limits\_{E}f$ is continuous respect to $E$ in metric $\delta$. Indeed, $f(E)-f(F) = \int f \cdot ( 1\_E - 1\_F )$, hence $|f(E)-f(F)| \leqslant \int |f| \cdot | 1\_E - 1\_F | = \int |f| \cdot 1\_{E \Delta F} = \int\limits\_{E\Delta F} |f|$. 


The last expression tends to $0$ if $|E\Delta F|$ tends to $0$ because of integrability of $f$, equivalently if $d(E,F)\to 0$. This remark shows that sets $\Sigma\_k$ are closed as an intersection of closed subsets of the space $(\Sigma,d)$. 


From Baire theorem we obtain that one of sets $\Sigma\_k$ has an interior point. Therefore, there exists a measurable set $E\_0$ and integer $k$ such that the inequality 
$ |f\_n(E)-f\_m(E) | \leqslant \varepsilon$ holds, whenever $|E\Delta E\_0| \leqslant \delta$ and $m,n\geqslant k$. We will show, that this inequality holds in fact for any set $E$, provided that its measure is sufficiency small. 


By identities $\mathbf{1}\_{E\cup E\_0} - \mathbf{1}\_{E\_0} = \mathbf{1}\_{E\cap E\_0^{c}}$ and $\mathbf{1}\_{E\_0}-\mathbf{1}\_{E\_0\setminus E} = \mathbf{1}\_{E\cap E\_0}$ we obtain for an arbitrary integrable $f$ 
$$f(E) = f(E \cap E\_0^{c}) + f(E\cap E\_0) = f(E\cup E\_0) - f(E\_0) + f(E\_0) - f(E\_0\setminus E)$$
If $|E| < \delta$, then all of sets $E\_0,E\_0\cup E, E\_0\setminus E$, belong to the ball $\{E:\ |E\Delta E\_0| < \delta \}$. Applying the last inequality to $f\_n-f\_m$ and $|E| < \delta$ we get
$$ |f\_n(E)-f\_m(E)| \leqslant 2\varepsilon \quad \textrm { if } |E|<\delta, \ n,m\geqslant k$$


Finally, observe that the finite family of integrable functions $f\_1,\ldots, f\_k$ is obviously locally uniformly integrable, i.e. $\sup\limits\_{i\leqslant k} |f\_i(E)| \to 0$ if $|E|\to 0$. Therefore, for sufficiency small $\delta'$ we have
$$|f\_i(E)|\leqslant \varepsilon \quad \textrm{ if } |E| < \delta', i\leqslant k$$ 


Gluing together two estimates that have been derived, we see that for some positive $\delta$ 


$$ \left|\int\limits\_{E} f\_n\right| \leqslant 3\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$


We have estimated integrals for functions $f\_n$, instead their modulus. It doesn't matter, however. Namely, applying the last estimate to the set $E\cap \{f\_n > 0\}$ and $E\cap \{f\_n < 0\}$ respectively (both contained in $E$ hence with a smaller measure), gives finally


$$ \int\limits\_{E} |f\_n| \leqslant 6\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$


What finished the proof.


