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Suppose $h_n$ is a sequence of non-negative functions in $L^2([0,1])$ converging weakly to $h$ (i.e., for every $g\in L^2([0,1])$ it holds $\int g\cdot h_n \,d\lambda \to \int g\cdot h\, d\lambda$).

Let $T_n\subseteq [0,1]$ be a measurable set for each $n$ such that $h_n$ vanishes almost everywhere on $T_n$. Suppose $T$ is a set such that the Lebesgue measure of the symmetric difference $T\Delta T_n := T\setminus T_n \cup T_n\setminus T$ goes to zero for $n\to \infty$.

Question: Does $h$ vanish almost everywhere on $T$?

Attempts: I tried to bound $\int_T h \, d\lambda < \varepsilon$ for arbitrary $\varepsilon >0$. Let $\varepsilon > 0$ and let $n_0$ such that for all $n\geq n_0$ we have $|\int_T h_n \, d\lambda - \int_T h \, d \lambda | < \varepsilon / 2$ (which exists by weak convergence). Then, we have $$ \int_T h \, d\lambda \leq \int_T h_n \, d\lambda + \varepsilon/2 $$ for all $n\geq n_0$. Now, I would want to use that $T$ is very close to $T_n$ and that $h_n$ vanishes on $T_n$. However, I cannot find a bound for the term $\int_{T\setminus T_n} h_n \, d\lambda$. Any ideas?

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    $\begingroup$ I don’t think bounding $\int_{T \setminus T_n} h_n \, d\lambda$ is possible. Say, if $h_n$ has support shrinking to the empty set, and $T_n$ is the complement of the support of $h_n$, then $T$ is $[0, 1]$ and the integral would just be the integral of $h_n$. $\endgroup$
    – David Gao
    Mar 14 at 22:07
  • $\begingroup$ Yes, I agree that this exact approach will likely not be successful. Do you believe there is a counter example? $\endgroup$
    – Michael
    Mar 14 at 22:26

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By passing to a subsequence, we may assume $\lambda(T \Delta T_n) < 2^{-n}$ for all $n$. Since the closures of a convex set under the weak topology and under the norm topology coincide, per Hahn-Banach theorem, we may choose $k_n \in \mathrm{co}(h_n, h_{n+1}, \cdots)$ such that $\|k_n - h\| < \frac{1}{n}$ for all $n$, where $\mathrm{co}$ indicates the convex hall. Then as $k_n \to h$ in norm, by passing to a subsequence if necessary, we may assume $k_n \to h$ a.e. Note that, as $k_n \in \mathrm{co}(h_n, h_{n+1}, \cdots)$, $k_n$ vanishes a.e. on $K_n = \cap_{m \geq n} T_m$. Since $\lambda(T \Delta T_m) < 2^{-m}$, we have $\lambda(T \Delta K_n) < 2^{-n} + 2^{-(n+1)} + \cdots = 2^{-n+1}$.

Clearly, as $h$ is the a.e. limit of $k_n$, $h$ vanishes a.e. on $\cup_{n \geq 1} \cap_{m \geq n} K_m$. But then,

$$T \setminus (\cup_{n \geq 1} \cap_{m \geq n} K_m) = \cap_{n \geq 1} (T \setminus [\cap_{m \geq n} K_m]) = \cap_{n \geq 1} (\cup_{m \geq n} [T \setminus K_m])$$

Since $\lambda(T \setminus K_m) < 2^{-m+1}$, we have $\lambda(\cup_{m \geq n} [T \setminus K_m]) < 2^{-n+2} \to 0$ as $n \to \infty$. Thus, $T \setminus (\cup_{n \geq 1} \cap_{m \geq n} K_m)$ is null, so $h$ vanishes a.e. on $T$.

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  • $\begingroup$ Thanks for the great answer! I needed some time to understand it, but I believe I got the gist of it now. One remark for the second paragraph: Instead of working with $\bigcup_n \bigcap_{m\geq n} K_m$, we can just as well consider the (equal) set $\bigcup_n K_n$ as $K_n=\bigcap_{m\geq n}K_m$ simplyfing the formulae slightly. $\endgroup$
    – Michael
    Mar 16 at 15:42
  • $\begingroup$ @Michael Ah, yes, you’re right. I forgot I was already defining $K_n$ as an infinite intersection and just wrote $\cup_n \cap_{m \geq n} K_m$ because that’s usually the set on which you can ensure an a.e. limit vanishes. In this specific case, yes, you already have $K_n = \cap_{m \geq n} K_m$ so you only need an infinite union. $\endgroup$
    – David Gao
    Mar 16 at 18:58

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