# Under what condition can converge in $L^1$ imply converge a.e.?

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!

• In short, convergence in $L^p$, $1\le p<+\infty$, is too weak for pointwise convergence, but one can find a subsequence that converges point-wise a.e. See more discussion here, here and here – A.Γ. Jul 24 '15 at 23:01
• BTW there is a nice review of different type of convergence by Terence Tao – A.Γ. Jul 24 '15 at 23:07
• Hint: Think about the condition $\sum_{n=1}^\infty c_n<\infty$. – user940 Jul 24 '15 at 23:15
• @A.G.Thanks! But my question is under what condition can converge in Lp imply converge a.e. In the problem is what condition on $c_n$. – Sherry Jul 24 '15 at 23:17
• @ByronSchmuland Thanks! Could you please give me more hint in proving this? Thanks! – Sherry Jul 24 '15 at 23:31

## 2 Answers

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 $x$.

• Thanks! Just to make sure, is the interchanging of summation and limit by Tonelli's theorem? – Sherry Jul 24 '15 at 23:39
• Well, monotone convergence is what I had in mind. – user940 Jul 24 '15 at 23:40
• (+1) All right, now I see. – Jack D'Aurizio Jul 24 '15 at 23:41
• @ByronSchmuland Thanks! Brilliant proof! – Sherry Jul 24 '15 at 23:42
• You're welcome. – user940 Jul 24 '15 at 23:47

One condition that works is $c_n=0$ for all $n\geq k$ for some $k>0$. This is because $f_n=0$ almost surely for each $n\geq k$, so let $A_n$ be the set on which $f_n\neq 0$. Then $\cup_{n}A_n$ has measure 0 by countable additivity, so $f_n\rightarrow 0$ almost surely.

• Thanks Alex. That is absolutely right for $c_n=0$ for all $n>k$. Could you explain why other conditions like $\sum _{n=0}^\infty c_n=0$ is not right? – Sherry Jul 24 '15 at 23:26