# Does the weak limit of a sequence in $L^2([0,1])$ vanish on the limit set of vanishing sets?

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?

• 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 $h_n$. Mar 14 at 22:07
• Yes, I agree that this exact approach will likely not be successful. Do you believe there is a counter example? Mar 14 at 22:26

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 $$T$$.
• 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. Mar 16 at 15:42
• @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. Mar 16 at 18:58