Let $m$ denote the Lebesgue measure on $[0,1]$. Let $f_n$ be a sequence of measurable functions defined on $[0,1]$ such that $0\le f_n\le M$ for each $n$. Moreover, assume that $\int_0^1f_n(x)dm(x)=1$ for all $n$. Consider a sequence $a_n$ such that $a_n\ge0$ for all $n$, and assume that $\sum_{n=1}^{\infty}a_n=\infty$.

Prove that there exists a subset $A$ of positive measure in $[0,1]$ such that $\sum_{n=1}^{\infty}a_nf_n(x)=\infty$ for each $x\in A$.

My attempt: Assume $\sum_{n=1}^{\infty}a_nf_n(x)<\infty$ almost everywhere on $[0,1]$. Since $a_nf_n\in L^{+}$, $\int_{[0,1]}\sum_{n=1}^{\infty}a_nf_n(x)=\sum_{n=1}^{\infty}\int_{[0,1]}a_nf_n(x)=\sum_{n=1}^{\infty}a_n\int_{[0,1]}f_n(x)=\sum_{n=1}^{\infty}a_n=\infty$. How can I get a contradiction?

Thanks in advance.

  • $\begingroup$ You can't. Because bounded almost everywhere doesn't imply convergent integral. I guess however you might have to make use of the uniform boundedness of $f_n$. $\endgroup$ – Vim Jun 27 '16 at 10:14

I finally solve it.

Assume $\sum_{n=1}^{\infty}a_nf_n(x)<\infty$ almost everywhere in $[0,1]$. Let $g_N(x)=\sum_{n=1}^Na_nf_n(x)$. $g_N$ is measurable. Let $g(x)=\sum_{n=1}^{\infty}a_nf_n(x)$. $g(x)$ is also measurable, and $g(x)<\infty$ a.e. Hence, $g_N(x)\to g(x)$ a.e. as $N\to\infty$. By Egoroff's theorem, for any $\epsilon>0$ exists $E\subset[0,1]$ such that $m(E)<\epsilon$ and $g_N(x)\to g(x)$ uniformly on $E^C$. By definition, $sup_{x\in E^C}|\sum_{n=N+1}^{\infty}a_nf_n(x)|\to0$. Thus, there exists $N_0\in\mathbb{N}$ such that for any $N>N_0$ for arbitrary $\delta>0$ we have $sup_{x\in E^C}\sum_{n=N+1}^{\infty}a_nf_n(x)<\delta$.

For such $N$, $\int_{E^C}\sum_{n=N+1}^{\infty}a_nf_n(x)\le m(E^C)\delta\le\delta<\infty$. Thus, $\sum_{n=N+1}^{\infty}a_n\int_{E^C}f_n(x)\le\delta$. Since $\int_{[0,1]}f_n=1$ and $\int_E f_n\le M\epsilon$. Hence, $\int_{E^C} f_n\ge 1-M\epsilon$. $\sum_{n=N+1}^{\infty}a_n(1-M\epsilon)\le\delta$. $\sum_{n=N+1}^{\infty}a_n\le\frac{\delta}{(1-M\epsilon)}$(take $\epsilon$ small enough). Then $\sum_{n=1}^{\infty}a_n\le\sum_{n=1}^{N}a_n+\frac{\delta}{(1-M\epsilon)}<\infty$. Contradiction.

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