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Let $(X,\mathcal{M},\mu)$ be a measure space such that $\mu(X)<\infty$. I want to show that if $(f_n)_{n \in \mathbb{N}}$ is a sequence of real-valued functions such that $|f_n (x)| \leq K$ for some real $K>0$ for every $n \in \mathbb{N}$ and $x \in X$, then $f_n \rightarrow f$ a.e. implies $f_n \rightarrow f$ in $L_1(X,\mu)$.

What we can do is use Ergoroff's Theorem that gives us, for every $\epsilon > 0$ a subset $A \subset X$ such that $\mu(A) < \epsilon$ and $f_n \rightarrow f$ uniformly in $X\setminus A$. So

$$\bigg|\int f_n d\mu - \int f d\mu \bigg| \leq \int|f_n - f|d\mu = \int_A |f_n - f| d\mu + \int_{X \setminus A} |f_n - f|d\mu \leq 2K\epsilon + \epsilon \mu(A) < 2K\epsilon + \epsilon^2 \ ,$$

because $f_n \rightarrow f$ in $X\setminus A$ and $f_n \rightarrow f$ in $A$ if $0< \mu(A) < \epsilon$.

I want to know if this argument is right, because the hypothesis of $\mu(X) < \infty$ is almost never used, except for the application of Ergoroff's Theorem above.

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  • $\begingroup$ The first inequality relies upon $\mu(X)<+\infty$ (when both $f_n$ and $f$ are uniformly bounded a.e.) or else the LHS need not be defined. Alternatively you may use dominated convergence. $\endgroup$ – H. H. Rugh Aug 12 '17 at 13:59
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Your proof is fine, except I don't see how you get the inequality $$ \int_{X\setminus A}|f_n-f|\,d\mu \leq \epsilon\mu(A). $$ Now you can obtain $$\int_{X\setminus A}|f_n-f|\,d\mu \leq \|(f_n-f)|_{X\setminus A}\|_\infty\mu(X)$$ instead, which will give you the result by taking $\|(f_n-f)|_{X\setminus A}\|_\infty$ small enough. You use the finiteness of the measure in Egoroff and also by implicitly assuming that $f_n$ and $f$ are in $L^1$. Here's a slightly different argument in case you're interested.

Since $f_n\to f$ a.e. and $|f_n|\leq K$, we get that $|f_n-f|\to0$ a.e. and $|f_n-f|\leq(|f_n|+|f|) \leq 2K$. Since the space is finite, $2K\in L^1$. Then the dominated convergence theorem implies $$ \int |f_n-f|\to 0. $$

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  • $\begingroup$ I didn't thought about using the dominated convergence theorem because I didn't knew if the $f_n$'s are in $L_1$, but I guess I need to assume this anyway. $\endgroup$ – B. Chinaski Aug 12 '17 at 14:12
  • $\begingroup$ The first inequality comes from the fact that $f_n \rightarrow f$ uniformly in $X \setminus A$, in which case I can bound the difference $|f_n -f|$ by $\epsilon$, isn't that right? $\endgroup$ – B. Chinaski Aug 12 '17 at 14:14
  • $\begingroup$ @B.Chinaski Yes you can, which is what I'm alluding to in the next line. $\endgroup$ – John Griffin Aug 12 '17 at 14:17

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