Definition (Pointwise convergence of sets): We say that a sequence $E_n$ of sets in $\mathbb{R}^d$ converges pointwise to another set $E$ in $\mathbb{R}^d$ if the indicator functions $1_{E_n}$ converge pointwise to the indicator function $1_E$.

The problem is to show that if $E_n$ are all Lebesgue measurable, and converge pointwise to $E$, then $E$ is also Lebesgue measurable.

Further, I need to show that $m(E_n)$ converges to $m(E)$ [where $m(\cdot)$ denotes the Lebesgue measure], if $E_n$ are all contained in another Lebesgue measurable set $F$ of finite measure. Also, I've to produce a counter example to show that the convergence may not hold without the assumption.

For the first part, I tried to write $E$ as some countable union/intersection of the $E_n$'s. But I cannot exploit the definition of pointwise convergence of sets well enough to defend my point. For the second part, probably I need to incorporate the monotone convergence theorem, but I don't have a clear-cut attack on the problem. Any help would be greatly appreciated!


A standard result is that the pointwise limit of a sequence of measurable functions is itself measurable, provided the ambiant space is complete (and $\mathbb{R}^d$ is complete). Since $$1_{E_n}(x)\rightarrow 1_{E}(x)$$ for every $x$ and $1_{E_n}$ is measurable, it follows that $1_E$ is also measurable, hence $E$ is a measurable set.

For the second part, we use the dominated convergence theorem. We know that $$1_{E_n}(x)\rightarrow 1_{E}(x)$$ for every $x$, while $$|1_{E_n}(x)|\leq 1_F(x)$$ which is an integrable function because $F$ has finite measure. It follows that $$m(E_n)=\int 1_{E_n}(x)dx\rightarrow\int 1_E(x)dx=m(E)$$ as $n\rightarrow\infty$.

  • $\begingroup$ The problem is that the book I'm following, has cited this second problem as the dominated convergence theorem (for Lebesgue measurable sets). So I think it'll be circular to use DCT here. Apparently, all that I have to attack this problem is the above definition and the monotone convergence theorem for Lebesgue measurable sets. I hope that you understand that I'm trying to avoid advanced machinery to avoid circularity. $\endgroup$ – user398842 Dec 14 '16 at 16:48
  • 2
    $\begingroup$ @aditi_ray I don't think Olivier's proof is circular, instead, I think his proof is totally right. In your comment, the only reason you give for the proof is circular is that the problem has a similar name to the theorem. I think the problem got this name just because it can be proved by DCT and has a similar form to DCT. Another point is it can be shown that the DCT, MCT and Fatou's theorem are all equivalent to each other, in the sense that one can prove another in any order. $\endgroup$ – Hua Dec 16 '16 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy