Real Analysis – Conditions for $L^1$ Convergence to Imply Almost Everywhere Convergence

almost-everywhereconvergence-divergencelebesgue-integralmeasure-theoryreal-analysis

Let $f_n$ be a sequence of Lebesgue measurable functions on $R^d$. Suppose you have an estimate of the form $\int_{R^d}\left|f_n\right|\le c_n$ where $c_n \downarrow 0$. Can you conclude that $f_n\to 0$ a.e.? If not, what additional conditions on ${c_n}$ would guarantee this?

My attempt:
I think we cannot conclude that $f_n\to 0$ a.e. For example $A_1=[0,1/2]$, $A_2=[1/2,1]$, $A_3=[0,1/4],\ldots,A_6=[3/4,1]$, $A_7=[0,1/8],\ldots$.
If $f_n$ is the indicator function of $A_n$, that is $f_n(x)=1$ if $x\in A_n$ and $f_n(x)=0$ else, then $f_n \to 0$ in all $L^p([0,1])$ because $\|f_n\|_p=\lambda(A_n)^{1/p}\to 0$ but there is no $x\in [0,1]$ with $f_n(x)\to 0$.

I have question in what additional conditions on ${c_n}$ would guarantee this?
Maybe $c_n$ strictly decreasing? However, I have trouble proving this. Could someone kindly help about this? Thanks!

Best Answer

Suppose that $$\int_{R^d} \sum_{n=1}^\infty |f_n(x)|\,dx= \sum_{n=1}^\infty \int_{R^d} |f_n(x)|\,dx \leq \sum_{n=1}^\infty c_n<\infty.$$

This shows that $\sum_{n=1}^\infty |f_n(x)|<\infty$ for almost every $x\in R^d$, and hence $|f_n(x)|\to0$ for such $x$.

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