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If $\limsup_{n\rightarrow\infty}\int\limits_{0}^{\epsilon}f_{n}(x)dx=0$ and $f_{n}(x)$ is a non negative function on the interval $[0,\epsilon]$ and $\epsilon$ is a non negative real number.

Then $\limsup_{n\rightarrow\infty}f_n(x)=0$ almost everywhere on the interval $[0,\epsilon]$?

I think it seems intuitively correct, but I don't know how to prove it. I would be appreciated if anyone can help me this. Thanks.

Siminore
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kapamax
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  • From your statement I can't understand if the limit relation must hold for any $\epsilon>0$ or just for a particular, fixed $\epsilon>0$. – Siminore Aug 07 '15 at 14:47

2 Answers2

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This would mean that $L^1$-convergence of a random variable implies almost sure convergence. However, this is false.

In order to see this, assume w.l.o.g. $\epsilon = 1$. Consider a sequence of indicators of intervals of the form $[k2^{-n}, (k + 1)2^{-n}]$ where k ranges from $0$ to $2^{-n}$. By these intervals I mean the sequence [0/1, 1/1], [0/2, 1/2], [1/2, 2/2], [0/4, 1/4], [1/4, 2/4], [2/4, 3/4], [3/4, 4/4], [0/5, 1/5], ...

Since the length of the intervals converges to 0, the corresponding indicators converge in the $L^1$-sense to 0. But pointwise these functions satisfy $\liminf f_n(x) = 0$ and $\limsup f_n(x) = 1$ for all $x \in [0, 1]$.

Dominik
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  • I suspect that $\epsilon$ is not for some $\epsilon$, but for every $\epsilon$. – Siminore Aug 07 '15 at 14:46
  • But the sequence of functions is only defined on $[0, \epsilon]$, so the phrasing of the claim would have to be changed significantly. I believe that my example remains valid if you define the sequence on $[0, 1]$ and expect the convergence of the integrals for all $\epsilon$. – Dominik Aug 07 '15 at 14:56
  • I agree with you. – Siminore Aug 07 '15 at 15:15
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Suppose $\{a_n\}$ is a sequence of nonnegative numbers and that $$\limsup_n a_n = 0.$$ By nonnegativity, we have $$0\le \liminf_n a_n.$$ Since $$0 \le \liminf_n a_n \le \limsup_n a_n = 0,$$ the limit of the $a_n$ must exist and be zero.

ncmathsadist
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