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This question arised when I was dealing with an old qual problem, and if this is true, I'll be done, but I'm not sure if it's true or not:

Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of non-negative Lebesgue integrable functions and suppose $$\int_{-1}^{1} f_n(x)\mathsf dx\stackrel{n\to\infty}\longrightarrow 0.$$ Then, can we conclude that $f_n(x)\stackrel{n\to\infty}\longrightarrow0$ for almost every $x\in [-1,1]$?

I'm motivated by the fact that when the integral of a non-negative function is zero, then it's 0 almost everywhere. I tried the negotiate the definition of almost everywhere convergence, but somehow couldn't get any contradiction yet, and really got stuck here. I would appreciate any kind of help.

Math1000
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vgmath
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    No, the standard counterexample is called the typewriter sequence. This question has certainly been asked many times before, but without a keyboard it's hard for me to search now. – Nate Eldredge Jul 30 '15 at 02:29

1 Answers1

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Consider the sequence defined by $$f_n = \begin{cases} \chi_{[0,1]} & n = 1\\ \chi_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} & 2^k \le n < 2^{k+1} \end{cases}$$ Then $\|f_n\|_1 = \frac1{2^k} \to 0$ but $f_n \not\to 1$ on any $x\in[0,1]$.

You can, however, prove that there exists an a.e. convergent subsequence with limit $0$ in this setting. Here you could chose $f_{2^k} = \chi_{[0,2^{-k}]} \to 0$ everywhere but at $0$.

Note that the sequence $f_n(x)$ is related to the binary expansion of $x$. If you take the first $k$ bits of $x$ (after the binary point) and convert that to a number $n$, $f_{2^k + n}(x) = 1$ and thus $f_{2^k + j}(x) = 0$ for $0\le j< 2^k$ and $j\ne n$. This means for fixed $x$, there will be at least one $0$ and one $1$ in $f_j(x)$ with $2^k \le j < 2^{k+1}$.

AlexR
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