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Let $(E, \mathcal{E}, \mu)$ be a measure space, and $(f_n)_{n\in\mathbb{N}}$ and $f$ be measurable functions $(E, \mathcal{E}, \mu)\longrightarrow (\mathbb{R}, \mathcal{B})$. The first part of Theorem 2.10 on page 20 of Stefan Grosskinsky's lecture notes says "assume that $\mu(E) < \infty$, then if $f_n \rightarrow f$ almost everywhere then $f_n \rightarrow f$ in measure". I don't quite understand why we need $\mu(E) < \infty$. The proof given in there does indeed use $\mu(E) < \infty$. However, I seem to be able to find another proof that doesn't need to use it.

I will first present (a slightly modified version of) the proof given by Prof. Grosskinsky below. Assume $f_n \rightarrow f$ almost everywhere and let $g_n = f_n - f$ for each $n$, we want to show that $g_n \rightarrow 0$ in measure. We need to show that for every $\epsilon > 0$, we have $\mu(|g_n| \geq \epsilon) \rightarrow 0$, or equivalently in the case $\mu(E) < \infty$, $\mu(|g_n| < \epsilon) \rightarrow \mu(E)$. Indeed, \begin{equation*} \begin{split} \mu(|g_n| < \epsilon) &\geq \mu\big(\bigcap_{m \geq n} \{|g_m| < \epsilon\}\big) \\ &\nearrow \mu\big(\bigcup_{N \in \mathbb{N}} \bigcap_{m > N} \{|g_m| < \epsilon\}\big) \\ &\geq \mu \big(\bigcap_{q \in \mathbb{Q},\, q > 0}\bigcup_{N \in \mathbb{N}} \bigcap_{m > N} \{|g_m| < \epsilon\}\big) \\ &= \mu(g_m \rightarrow 0) \\ &= \mu(E). \end{split} \end{equation*}

I understand the proof above, but below I will present what seems to be a proof without resorting to $\mu(E) < \infty$. This proof directly shows, for all $\epsilon > 0$, that $\mu(|g_n| \geq \epsilon) \rightarrow 0$, as opposed to the proof above where we showed $\mu(|g_n| < \epsilon) \rightarrow \mu(E)$. \begin{equation*} \begin{split} \mu(|g_n| \geq \epsilon) & \leq \mu\big(\bigcup_{m \geq n} \{|g_m| \geq \epsilon\}\big)\\ &\searrow \mu \big(\bigcap_{N \in \mathbb{N}} \bigcup_{m>N} \{|g_m| \geq \epsilon\}\big)\\ &\leq \mu\big(\bigcup_{q\in\mathbb{Q},\, q>0} \bigcap_{N \in \mathbb{N}} \bigcup_{m>N} \{|g_m| \geq \epsilon\}\big)\\ &= \mu(g_m \nrightarrow 0)\\ &= 0. \end{split} \end{equation*} This 'proof' doesn't resort to $\mu(E) < \infty$, so it seems to me. Am I doing something wrong? Thank you in anticipation for your help!!!

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    $\begingroup$ To conclude that $\mu \bigl(\cap_n A_n\bigr)=\lim_n \mu(A_n)$ for $A_1\supset A_2\supset A_3\supset \cdots$, you need $\mu(A_i)<\infty$ for some $i$. Consider the sets $A_n=[n,\infty)$. (Try your proof applied to $g_n=\chi_{[n,\infty)}$.) $\endgroup$ – David Mitra Feb 18 '14 at 22:36
  • $\begingroup$ David thank you for your answer! Forgive me for a silly question but what is $\chi_{[n, \infty)}$??? $\endgroup$ – Ray Feb 19 '14 at 14:11
  • $\begingroup$ The characteristic function of $[n,\infty)$. $\endgroup$ – David Mitra Feb 19 '14 at 14:16
  • $\begingroup$ Ah yes of course! Thank you very much! Would you like to re-post your first comment as an answer so I can mark it as correct? $\endgroup$ – Ray Feb 19 '14 at 15:17
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You're using the following result in your proof attempt:


Suppose $A_1$, $A_2$, $A_3$, $\ldots$ are measurable sets with $A_1\supset A_2\supset A_3\supset\cdots$. If $\mu(A_i)<\infty$ for some $i$, then $\mu\Bigl(\,\bigcap\limits_{k=1}^\infty A_k\Bigr)=\lim\limits_{k\rightarrow\infty} \mu(A_k)$.


The hypothesis that some $A_i$ has finite measure is necessary. Consider the sets $A_k=[k,\infty)$, $k\ge1$.

(It should be clear why things don't work in your proof if you consider the functions $\chi_{[n,\infty)}$ ($\chi_E$ is the characteristic function on $E$). The sequence $(\chi_{[n,\infty)})$ converges to $0$ pointwise everywhere in $\Bbb R$, but does not converge in measure.)

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