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I wrote up the following proof for the lemma, please check if I made any mistakes, thank you!

Statement: Suppose that $f_n,f : X\rightarrow \mathbb{R}$ are measurable functions such that $\sum_{n=1}^{\infty} ||f_n - f||_{L^1} < \infty$, then $f_n \rightarrow f$ almost uniformly; which also implies $f_n \rightarrow f$ point wise almost everywhere.

Proof: Using Markov's inequality, for each $\lambda>0$ $$\lambda*\mu\left(\{x\in X :|f_n(x)-f(x)|\geq \lambda\}\right) \leq ||f_n-f||_{L^1} .$$ Summing over $n$ and dividing by $\lambda$ we have $$\sum_{n=1}^\infty \mu\left(\{x\in X :|f_n(x)-f(x)|\geq \lambda \}\right) <\infty.$$ For each $\epsilon > 0$, there exists $N\in\mathbb{N}$ such that $$\sum_{n=N}^\infty \mu \left(\{x\in X :|f_n(x)-f(x)|\geq \lambda\}\right) <\epsilon,$$ or $$\mu\left(\cup_{n=N}^{\infty}\{x\in X :|f_n(x)-f(x)|\geq \lambda\}\right) <\epsilon.$$ And outside of this exceptional set, $f_n \rightarrow f$ uniformly since $\lambda$ is arbitrary.

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1 Answer 1

Summing the first inequality over $n$, I get $$\lambda \sum_{n=1}^\infty \frac{1}{2^n} \mu\left(\{x\in X :|f_n(x)-f(x)|\geq \frac{\lambda}{2^n} \}\right) \leq ||f_n-f||_{L^1} $$ with the factor of $1/2^n$ hanging in there. You can't just sum it as a geometric series, it's attached to a quantity dependent on $n$. In a nutshell, $a_1b_1+a_2b_2 \ne (a_1+a_2)(b_1+b_2)$.

To fix this, use $\lambda$ without $2^n$: $$\lambda \sum_{n=1}^\infty \mu\left(\{x\in X :|f_n(x)-f(x)|\geq \lambda \}\right) \leq ||f_n-f||_{L^1} $$ This directly gives the conclusion $$\sum_{n=1}^\infty \mu \left(\{x\in X :|f_n(x)-f(x)|\geq \lambda\}\right) <\infty$$ and the proof proceeds as you wrote.

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Yes, Thank you. I edited my proof. –  Xiao Mar 20 '14 at 9:52

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