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Let $E$ be a measurable set, $(f_n)$ a sequence of real valued measurable functions on $E$ and $f$ a real valued measurable function on $E$. It is required to prove that if $(f_n)$ converges to $f$ almost uniformly then $(f_n)$ converges to $f$ in measure. The following is my proof.

Let $\epsilon>0$. Suppose $(f_n)$ converges to $f$ almost uniformly. Then there exists $F\subseteq E$ such that $m(F)<\epsilon$ and $f_n$ converges uniformly to $f$ on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n\geq N$ and $x\in E\setminus F,$ $|f_n(x)-f(x)|<\epsilon.$

But\begin{align} \{x\in E:|f_n(x)-f(x)|\geq\epsilon\}=\{x\in F:|f_n(x)-f(x)|\geq\epsilon\}\cup\{x\in E\setminus F:|f_n(x)-f(x)|\geq\epsilon\}.\end{align}

Let $n\geq N$. Then $m(\{x\in F:|f_n(x)-f(x)|\geq\epsilon\})<\epsilon$ and

$m(\{x\in E\setminus F:|f_n(x)-f(x)|\geq\epsilon\})=0$.

Therefore for each $n\geq\mathbb{N}$, $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})<\epsilon$.

Hence $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})=0$ as $n\to\infty$ and the proof is complete.

Is this proof alright? Thanks.

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  • $\begingroup$ Yes, your proof is essentially correct. I posted an answer with more details. Let me know if you have any question. $\endgroup$ – Ramiro Jun 6 '17 at 23:53
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Yes, your proof is essentially correct. Here it is with some small improvements.

Let $\epsilon>0$. Suppose $(f_n)$ converges to $f$ almost uniformly. Then there exists $F\subseteq E$ such that $m(F)<\epsilon$ and $f_n$ converges uniformly to $f$ on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n\geq N$ and $x\in E\setminus F,$ $|f_n(x)-f(x)|<\epsilon.$

It means, for $n\geq N$, $$ E\setminus F \subseteq \{x\in E:|f_n(x)-f(x)|<\epsilon\} $$

So, for $n\geq N$, $$ \{x\in E:|f_n(x)-f(x)|\geq\epsilon\} \subseteq F$$ and so we have, for $n\geq N$, $$ m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})\leqslant m(F)<\epsilon$$

Hence $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})=0$ as $n\to\infty$ and the proof is complete.

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