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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: Let $\|\cdot\|$ denote $L^1$ norm, given $\sum_n \|f_n - f\|<\infty$, choose a sequences $\{c_n\}$ such that $\{c_n\}$ increases to $\infty$ and yet
$$\sum_n c_n \|f_n - f\|< \infty.$$

Using Chebychev ineq, we have $$\frac{1}{c_n} \mu(\{x\in X : |f_n - f|\geq \frac{1}{c_n}\}) \leq \|f_{n} - f\| $$ $$ \mu(\{x\in X : |f_n - f|\geq \frac{1}{c_n}\}) \leq c_n \|f_{n} - f\|.$$

For each $\epsilon > 0$, there exists a $N$ such that $$\sum_{n=N}^\infty c_n \|f_{n} - f\| < \epsilon,$$ Define $$A:=\bigcup_{n=N}^{\infty} \{x\in X : |f_{n} - f|\geq \frac{1}{c_n}\}$$ then $$\mu(A) \leq \sum_{n=N}^\infty \mu(\{x\in X : |f_n - f|\geq \frac{1}{c_n}\}) \leq \sum_{n=N}^\infty c_n \|f_{n} - f\| \leq \epsilon$$ and $$\lim_{n\rightarrow \infty} \sup_{x\not\in A}\|f_{n} - f\|_\infty \leq \frac{1}{c_n} = 0$$ which means $f_n$ converges uniformly to $f$ outside of $A$.

For the existence of $c_n$, see here.

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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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  • $\begingroup$ Yes, Thank you. I edited my proof. $\endgroup$ – Xiao Mar 20 '14 at 9:52
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I think that this proof is incorrect. You need to show that for any $\varepsilon >0$, there exists $A\subset X$ so that $sup_{x\notin A}{|f_n(x)-f(x)|\to 0}$.

You proved that for any $\varepsilon>0$ and $\lambda >0$, there exists $A\subset X$ so that $|f_n(x)-f(x)|<\lambda$ for large $n$ and $x\notin A$.

$A$ here is dependent on $\lambda$ which is forbiden.

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  • $\begingroup$ Yes, it was incorrect, I kind of forgot about this one. Updated with new proof. $\endgroup$ – Xiao Feb 2 '15 at 20:09

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