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Suppose $\{u_n\}_{n=1}^{\infty}$ is an orhtonormal basis in $L^2[0,1]$, prove that $\sum_{n=1}^{\infty}|u_n(x)|^2=\infty$ for almost every $x\in [0,1]$.

Any hint on this problem?

I tried to prove the set $Y$ has measure zero, where $Y=\{x\in [0,1]: \sum_{n=1}^{\infty}|u_n(x)|^2<\infty\}$. Then by decomposition, WLOG, we only need to show that $Y_k=\{x\in [0,1]: k\leq \sum_{n=1}^{\infty}|u_n(x)|^2 has measure zero for any $k\in \mathbb{Z}^+$, then I tried to prove by contradiction. But I can only show that $m(Y_k)\geq \frac{1}{k+1}$, then I have no idea how to proceed the proof.

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    I suspect your argument for proving $m(Y_k)\le m(Y_k)^2(k+1)$ also works when $Y_k$ is replaced with any of its measurable subsets(just like my answer). Then you can complete your proof.2012-12-03

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Given $k\in\mathbb{N}$, let $X_k=\{x\in[0,1]:\sum_{n=1}^\infty|u_n(x)|^2\le k\}.$ To prove your conclusion, it suffices to show that for every $k\in\mathbb{N}$, $m(X_k)=0$. To prove $m(E_k)=0$, it suffices to show that if $E$ is a measurable subset of $X_k$ with $km(E)<1$, then $m(E)=0$.

Fixing such a set $E$, since $\{u_n\}_{n=1}^\infty$ is an orthonormal basis of $L^2[0,1]$, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n$ converges to $\chi_E$ in $L^2[0,1]$. Therefore, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E$ also converges to $\chi_E$ in $L^2[0,1]$. Since $E\subset X_k$, by Cauchy-Schwarz inequality, for every $x,y\in E$, $\sum_{n=1}^\infty|u_n(x)\bar{u}_n(y)|\le k$. It follows that
$\chi_E=|\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E|\le km(E)\chi_E<\chi_E \quad\mbox{a.e. on } E,$ which implies that $m(E)=0$.

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    @ougao: You are welcome!2012-12-03