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Egorov's Theorem: Let $(X,M,\mu)$ be a finite measure space and $f_n$ a sequence of measurable functions on $X$ that converges pointwise a.e. on $X$ to a function $f$ that is finite a.e. on $X$. Then for each $\epsilon>0$, there is a measurable subset $X_{\epsilon}$ of $X$ for which $f_n→f$ uniformly on $X_{\epsilon}$ and $\mu(X\backslash X_\epsilon)<\epsilon$.

Proof:
Let $X_0$ be the set on which $f_n \to f$, so $\mu(X \backslash X_0)=0$.
Let $m \in \Bbb N$. For every $x \in X_0$ we can find $n \in \Bbb N : |f_k(x)-f(x)|< \frac 1m\ \forall k \gt n$. Now define $$A_n^m = \{x\in X_0 : |f_n(x)-f(x)|< \frac 1m\},$$ and set $X_n^m=\bigcap_{k=n}^{\infty}A_k^m$. Since the $A_n^m$ are measurable and $X_n^m$ is the countable intersection of measurable sets, the $X_n^m$ are measurable. The $X_n^m$ are ascending, since as $n$ grows one is intersecting fewer and fewer of the $A_n^m$.

And so on...

And question is: How can I show that $\bigcup_{n=1}^{\infty}X_n^m=X_0$? It is taken as obvious in books, but if my teacher asks me about it I will not have an answer.

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    $\begingroup$ Try to describe $X_n^m$ in words: it is exactly the set of elements $x\in X_0$ such that $|f_k(x)-f(x)|<\frac{1}{m}$ for any $k\ge n$. But as you said for any $x\in X_0$ (and any fixed $m$) there is some $n$ such that this holds! $\endgroup$
    – Mizar
    Commented Dec 28, 2014 at 20:15
  • $\begingroup$ Thanks a lot! Understood now. $\endgroup$
    – Ilya
    Commented Dec 28, 2014 at 20:48

2 Answers 2

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Try to describe $X^m_n$ in words:
$X^m_n$ consists exactly of those elements $x\in X_0$ such that $|f_k(x)−f(x)|<\frac{1}{m}$ for any $k\ge n$.
But as you said for any $x\in X_0$ (and any fixed $m$) there is some $n$ such that this holds!

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An arguably more efficient way to see this is to keep in mind the close relation between the logical quantifiers $\forall,\exists$ and the set operations $\cap,\cup$, respectively. In this regard, an argument showing $\bigcup_n X_n^m=X_0$ might go like this:

Let $x\in X_0$. Since $f_n(x)\to f(x)$, $\forall\epsilon>0,\exists n\geq1,\forall k\geq n: |f_n(x)-f(x)|<\epsilon$. Let $m\in\mathbb{N}$. Then

\begin{align} &\exists n\geq1,\forall k\geq n:|f_n(x)-f(x)|<\epsilon:=1/m\\ &\implies\exists n\geq1,\forall k\geq n:x\in A_n^m\\ &\implies\exists n\geq1, x\in \bigcap_{k\geq n} A_n^m=X_n^m\\ &\implies x\in \bigcup_{n\geq1} X_n^m \left(= \bigcup_{n\geq1} \bigcap_{k\geq n} A_n^m = \liminf_n A_n^m \right). \end{align}

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