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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?

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2 Answers 2

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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.

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Very nice! $\mbox{}$ –  Nate Eldredge Mar 14 '11 at 15:58
    
An interesting exercise :-) However, my professor told me, when I showed him this proof, that it is possible to give a short proof without the Baire theorem, using Schur property of the sequences space $l^{1}$ –  Maciej S. Mar 14 '11 at 17:03
    
Very beautiful proof. –  Beni Bogosel Mar 15 '11 at 6:04

By the way, in a posted solution we used only following properties of functionals $f(E)=\int\limits_{E} f$:

  1. Continuous respect to $|E|$ (Lebesgue measure)
  2. Additive (on disjoint sets)
  3. Finite

Therefore, our problem can be reformulated in a following way:

Suppose, that $\mu_n$ is a sequence of finite measures, absolutely continuous w.r.t. the Lebesgue measure. Then $\sup\limits_{n} \ |\mu_n|(E) \to 0$ if $|E|\to 0$.

So we have proved Vitali-Hahn-Saks theorem :-)

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