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.
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).
