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Let $(f_n)_{n\in \mathbb{N}}$ be a sequence of nonnegative measurable functions in $L_1[0,1]$. Assume that $$f_n \to f, \text{ a.e.}$$ and $$\int f_n h \to \int g h,\, \forall h \in C[0,1].$$

Question: Do we have $$f = g, \text{ a.e.?}$$

Remark: Of course, it is a standard result that if we allow $h$ to be any bounded measurable function, then the almost sure limit coincides with the weak limit.

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    $\begingroup$ This does not follow. Take $f_n$ very large on a very small set $A_n$ so that $\int f_n=1$; this set should be spread out over $(0,1)$. Then $f=0$ (definitely if $|A_n|$ is summable), $g=1$. $\endgroup$ – Christian Remling Jan 3 '15 at 21:07
  • $\begingroup$ Indeed, this is a nice counter-example. $\endgroup$ – Yanqi QIU Jan 3 '15 at 21:30
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Exploiting Christian Remling's idea, we can take the following example: $$f_n:=b_n\mathbb 1\left(\bigcup_{j=1}^n\left(j /n-a_n,j/n+a_n\right) \right),$$ where the sequence $(a_n)$ is such that the series $\sum_n na_n $ converges and $na_nb_n=1$ for each $n$. Since for a continuous function $h$, the inequality $$\left|\int f_nh\mathrm dx-\frac 1n\sum_{j=1}^nh(j/n)\right| \leqslant nb_na_n \sup_{\substack{ s,t\in[0,1]\\ |s-t|\leqslant a_n } }|h(t)-h(s)|= \sup_{\substack{ s,t\in[0,1]\\ |s-t|\leqslant a_n } }|h(t)-h(s)|, $$ hence we have $$\lim_{n\to \infty}\int f_nh\mathrm dx=\int h\mathrm dx.$$

Since $\sum_n\lambda\{x\mid f_n(x)\neq 0\}\leqslant \sum_nna_n$, we have $f_n\to 0$ almost everywhere (by the Borel-Cantelli lemma).

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