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I am dealing with a problem at the moment where the hypothesis can be restated as $\int_{0}^{1}f_{n}\;dx\to0$ and $f_{n}\geq0$. Under these conditions, I want to conclude that $\lim f_{n}$ exists and is $0$ for almost every $x$ in $[0,1]$. Without boundedness or monotonicity, the usual convergence theorems are not immedaitely applicable. Also, if the limit of integrals converged to anything other than $0$, or if the $f_{n}$ were allowed to be signed, or if the set of integration had infinite measure, well known counter-example(s) would be available. It seems that the finite measure of $[0,1]$, the nonnegativity of $f_{n}$, and the assumption $\int_{0}^{1}f_{n}\;dx\to0$ should force $f_{n}\to0$ by appealing (in some manner) the well known fact that for $g\geq0$ measurable, $\int_{0}^{1}g\;dx=0$ if and only if $g=0$ a.e. $x\in[0,1]$. The only way out of this is if the $f_{n}$ spiked on sets of small measure; but as $n\to\infty$, these sets where the $f_{n}$ spike must become correspondingly smaller (in measure) since we have $\int_{0}^{1}f_{n}\;dx<\epsilon$ for $n$ large. In the limit, these "spike sets" should yield to a null set, thus proving the claim.

The question I am referring to is here Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant

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Sargera
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1 Answers1

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The canonical counterexample is to take the indicator functions of $[0,1]$; $[0,1/2]$,$[1/2,1]$, $[0,1/4]$, $[1/4,1/2]$, $[1/2,3/4]$, $[3/4,1]$ &c.

(If the pattern is not evident: break $[0,1]$ into $2^k$ intervals let $f_n$ be the sequence of indicator functions of each $2^k$ intervals obtained at each step from left to right, let $k=0,1,2,\ldots$)

In this case $\liminf f_n=0$ while $\limsup f_n=1$ so the sequence of functions converges nowhere, yet $$\int_0^1 f_n\to 0$$

Pedro
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  • @T.Bongers Yes. ${}{}$ – Pedro Sep 06 '13 at 03:18
  • Ah. Typewriter sequence.

    So what if we add the auxiliary condition that $\int_{0}^{1}f_{n};dx\to0$ "rapidly" in the sense that $n\int_{0}^{1}f_{n};dx\to0$ as $n\to\infty$? Not that the counter-example exploited possible unboundedness of $f_{n}$, but also assume $0\leq f_{n}\leq M$ along with the rapid convergence condition. This is the condition of the convergence in the question I linked in the original post.

     – Sargera Sep 06 '13 at 03:29
  • @Sargera But this is also a counterexample to that, is it not? We have that $\int_0^1 f_n=2^{-k_n}$ for $k_n$ equal to $0,1,1,2,2,2,2,3,3,\ldots$ so this certainly does it. – Pedro Sep 06 '13 at 03:34
  • You're right. I think I need to take a second and rethink, since actually in my comment I really want $\lim nf_{n}$ to converge to $0$ a.e. $x$, not just $f_{n}$. – Sargera Sep 06 '13 at 03:35