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Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Let $f_n$ be a sequence of measurable real-valued functions such that $f_n$ converges pointwise a.e. to a real-valued function $f$.

Show that for each point in $X$, the following limit holds

$$\lim_{n\to\infty} f_n \chi_{\displaystyle\{x\in X: |f_n(x)|\leq \alpha\}}=f \chi_{\displaystyle\ \{x\in X: |f(x)|\leq \alpha\}}$$

With exception of countably many values of $\alpha$.

I have not idea how to do it, I tried it 3 hours. I tried with Fatou's, LDCT, LMCT, Borel-Cantelli etc...

Edit: Etienne in answers noted that the conclusion does not hold if the exeption is in finitely many values of $\alpha$.

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What kind of convergence are you trying to prove? Pointwise a.e., convergence in measure? –  Alex Becker Jun 5 '13 at 19:39
    
@Alex Sorry I mean pointwise. –  Gastón Burrull Jun 5 '13 at 20:08
    
What is the source of this problem? –  Nate Eldredge Jun 6 '13 at 3:37
    
@Nate I don't know, I saw it in an exam. –  Gastón Burrull Jun 6 '13 at 4:35

1 Answer 1

up vote 2 down vote accepted

If the limit is understood as "pointwise a.e.", it seems that the conclusion does not necessarily hold. Assume that the space $X$ can be written as $X=\bigcup_{k\in\mathbb N} A_k$, where the $A_k$'s are (measurable,) pairwise disjoint with $\mu(A_k)>0$ (for example, take $X=[0,1]$ withLebesgue measure). Let $(\alpha_k)$ be any sequence of pairwise distinct positive numbers, and define a measurable function $f$ on $X$ by $f(x)=\alpha_k$ if $x\in A_k$. Finally, define $f_n(x)=f(x)+\frac1n\cdot$ Then $f_n(x)\to f(x)$ everywhere, but for each fixed $k$ every $x\in A_k$ we have $\chi_{\{\vert f_n\vert\leq \alpha_k\}}(x)=0$ for all $n$ (since $f_n(x)=\alpha_k+\frac1n$) whereas $\chi_{\{\vert f\vert\leq \alpha_k\}}(x)=1$ (since $f(x)=\alpha_k$); hence, for any $k\in\mathbb N$, the above limit does not hold with $\alpha=\alpha_k$.

On the other hand, the conclusion does hold (with pointwise a.e. convergence) if "finitely many values of $\alpha$" is replaced with "countably many values of $\alpha$". The key point is that the set $A$ of all $\alpha\geq 0$ such that $\{ x\in X;\; \vert f(x)\vert=\alpha\}$ has positive measure is countable, because the measure $\mu$ is finite ($\sigma$-finite would be enough). If $\alpha\not\in A$, then $\vert f(x)\vert \neq\alpha$ almost everywhere (by the definition of $A$), so that $\vert f(x)\vert$ is a point of continuity of the function $\chi_{[0,\alpha]}$. Since $\vert f_n(x)\vert\to\vert f(x)\vert$ a.e., it follows that $\chi_{[0,\alpha]}(\vert f_n(x)\vert)\to \chi_{[0,\alpha]} (\vert f(x)\vert)$ almost everywhere for any $\alpha$ not in the countable "exceptional" set $A$, and this shows that the above limit holds true for these $\alpha$.

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How to deduce "$A$ numerable" from "$\mu$ is $\sigma$-finite or finite"? –  Gastón Burrull Jun 6 '13 at 1:42
    
The sets $E_\alpha=\{ x\in X;\; \vert f(x)\vert=\alpha\}$ are pairwise disjoint, and in a $\sigma$-finite measure space you cannot find an uncountable family of pairwise disjoint sets with positive measure. –  Etienne Jun 6 '13 at 5:54
    
Sorry but I still don't get it. Why the existence of uncountable disjoint family of positive measurable sets contradicts the fact that $\mu(X)<\infty$? I really can't see why. –  Gastón Burrull Jun 6 '13 at 18:45
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OK, here is a detailed proof. Assume the measure space is finite. Let $(E_i)_{i\in I}$ be a family of pairwise disjoint measurable sets with positive measure. We want to show that the index set $I$ is necessarily countable. For each $n\in\mathbb N$, put $I_n=\{ i\in I;\; \mu(E_i)\geq \frac 1n\}$. Then $I=\bigcup_{n\in\mathbb N} I_n$ since $\mu(E_i)>0$ for all $i$. So it is enough to show that each $I_n$ is finite. But this is clear since $$\sum_{i\in I_n} \mu(E_i)=\mu\left(\bigcup_{i\in I_n}E_i\right)<\infty$$ and $\mu(E_i)\geq \frac1n$ for all $i\in I_n$. –  Etienne Jun 6 '13 at 18:53
    
Thanks I really appreciate your help. –  Gastón Burrull Jun 6 '13 at 19:09

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