# Dominated Convergence Theorem for Sets

This was an interesting question, which gives the analog of the better known Dominated Convergence Theorem for Lebesgue integrable functions.

Suppose $E_{n}\to E$ pointwise (e.g. the indicator functions $\chi_{E_{n}}\to\chi_{E}$ pointwise) and that $E_{n}\subset F$ for all $n$ where $m(F)<\infty$. (Note, all sets in this exercise are assumed to be Lebesgue measurable on $\mathbb{R}^{d}$).

Then show that $E$ is Lebesgue measurable, that $\lim_{n\to\infty}m(E_{n})=m(\lim_{n\to\infty}E_{n})=m(E)$, and finally that the hypothesis $m(F)<\infty$ cannot be dropped, even when $m(E_{n})<\alpha$ for some uniform bound.

The first part is easy, since one just uses the definition of $\lim\sup$ or $\lim\inf$ to express $E$ as a countable intersection/union of Lebesgue measurable sets. The third part is also clear, since one can just translate well-known examples for Riemann integrals into their set analogs. For example, take the $E_{n}$ to be the regions in $\mathbb{R}^{2}$ bounded by $n\cdot\chi_{(0,\frac{1}{n})}$ and the x axis; then one has $\lim_{n\to\infty}m(E_{n})=1$, yet $m(\lim_{n\to\infty}E_{n})=0$. The obvious problem is that the sets $E_{n}$ escape to "vertical infinity" near the origin; the dominated convergence shuts down this possibility. But the example is helpful, since while the process of commuting limits (even in the Lebesgue theory; although at least in the Lebesgue theory, measurable sets and functions are closed under limits, unlike the Riemann/Jordan theory) is not generally permissible, the intuition that they should commute remains valid under the general blanket hypotheses of the convergence theorems, which is where you intuition comes from anyway.

Anyway, here's my attempt to the solution of part (2). I have a feeling that it should be simpler than this massive computation; but even if it's not, I want to verify that mine is correct, since I haven't had much experience working with $\lim\sup$ and $\lim\inf$ operations of sets/sequences.

Since $E_{n}\to E$, we have the equality of $\lim\sup$ and $\lim\inf$ of the sequence, and so $$E=\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}=\bigcap_{j=1}^{\infty}\bigcup_{k=j}^{\infty}E_{k}.$$ Thus, $E$ is the countable intersection and union of Lebesgue measurable sets, and so is itself a Lebesgue measurable set.

The following computation \begin{align*} m(E) &=m\left(\lim_{n\to\infty}E_{n}\right)\\ &=m\left(\lim\inf\limits_{n\to\infty}E_{n}\right)\\ &=m\left(\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_{k}\right)\\ &=m\left(\lim_{n\to\infty}\bigcap_{k=n}^{\infty}E_{k}\right)\\ &=\lim_{n\to\infty}m\left(\bigcap_{k=n}^{\infty}E_{k}\right)\\ &=\lim_{n\to\infty}m\left(\inf_{k\geq n}E_{k}\right)\\ &\leq\lim_{n\to\infty}\inf_{k\geq n}m(E_{k})\\ &=\lim\inf\limits_{n\to\infty}m(E_{n})\\ &\leq\lim\sup\limits_{n\to\infty}m(E_{n})\\ &=\lim_{n\to\infty}\sup_{k\geq n}m(E_{k})\\ &\leq\lim_{n\to\infty}m\left(\sup_{k\geq n}E_{k}\right)\\ &=\lim_{n\to\infty}m\left(\bigcup_{k=n}^{\infty}E_{k}\right)\\ &=m\left(\lim_{n\to\infty}\bigcup_{k=n}^{\infty}E{k}\right)\\ &=m\left(\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_{k}\right)\\ &=m\left(\lim\sup\limits_{n\to\infty}E_{n}\right)\\ &=m\left(\lim_{n\to\infty}E_{n}\right)\\ &=m(E) \end{align*} The computations are justified by the following observations. First, $E_{n}\to E$ implies that $\lim\sup_{n\to\infty}E_{n}=\lim\inf_{n\to\infty}E_{n}=\lim_{n\to\infty}E_{n}=E$. By definition $\lim\inf_{n\to\infty}=\bigcup_{n=1}^{\infty}\bigcap_{k\geq n}^{\infty}E_{k}=\lim_{n\to\infty}\bigcap_{k=n}^{\infty}E_{k}$, and if we define $G_{n}=\bigcap_{k\geq n}^{\infty}E_{k}$, we see that $G_{n}\subset G_{n+1}$ for all $n$. Hence, the $G_{n}$ form an upward monotonic sequence of sets and so we may apply upward monotonic convergence theorem to establish the commutation of the first limit. It is helpful to use the technically incorrect notation $G_{n}=\inf_{k\geq n}E_{k}$ (where the $\inf$ is understood to be with respect to set inclusion). Then it becomes clear that $G_{n}\subset E_{k}$ for all $n\leq k$, so from monotonicity we have $m(G_{n})\leq m(E_{k})$ for all $k\geq n$, and so the above inequality holds. The remaining statements follow from the definition of $\lim\inf_{n\to\infty}$ for a sequence of real numbers. Justifications of the $\lim\sup$ computations are essentially the same as for the $\lim\inf$ case (though the computations are backwards since I wanted to combine everything into one inequality); except that we apply the downward monotone convergence theorem to the sequence defined by $G_{n}=\bigcup_{k\geq n}^{\infty}E_{k}$ to establish the commutation of the final limit. This is valid since $G_{n}\supset G_{n+1}$ for all $n$, and since $E_{k}\subset F$ and $m(E_{k})<m(F)<\infty$ hold for all $k$, we see that each $G_{k}$ is bounded, and so any union of the $G_{k}$ is bounded with finite measure. The inequality is of course obtained from monotonicity and noting that $G_{n}\supset E_{k}$ for all $k\geq n$. Putting this together then gives us the inequality $$\lim\inf_{n\to\infty}m(E_{n})\leq m\left(\lim_{n\to\infty}E_{n}\right)\leq\lim\sup\limits_{n\to\infty}m(E_{n}).$$ But since $E_{n}$ converges, it is clear that we have an equality, and so $$\lim\inf_{n\to\infty}m(E_{n})=\lim\sup\limits_{n\to\infty}m(E_{n})=\lim_{n\to\infty}m(E_{n})=m\left(\lim_{n\to\infty}E_{n}\right)=m(E)$$ as required.

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In the beginning you also wanted to write '$E_n\subseteq F$' instead of the other way around, no? – Berci Oct 23 '12 at 9:25

It all seems correct (after I could resolve the typo '$E_n\supset F$' that it actually means '$E_n\subseteq F$' for all $n$), and that the thin lanes are $[0,\frac1n]\times[0,n]$ rectangles perhaps..

Do you still have some question about it?

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I fixed a couple of the typos. There was an $=$ sign instead of a $\leq$ sign on line 11 of the computation. Indeed, $E\subset F$, since otherwise the sequence is dominating $F$! (; I actually meant $n\cdot\chi_{(0,\frac{1}{n})}$, if that is what you were referring to. – Taylor Martin Oct 23 '12 at 9:45
I suppose there's no more question then. I was just a little suspicious of my reasoning, particularly at the end with "But since $E_{n}$ converges, it is clear that we have an equality, and so$\ldots$"; I'm not sure how to show that rigorously, so I'm just afraid of a looming counter-example. Thank you for reviewing my solution though! – Taylor Martin Oct 23 '12 at 9:48
I do want to make sure of ONE last thing (I'll re-accept the answer)... – Taylor Martin Oct 23 '12 at 10:06
(Ignore above; couldn't edit after 5 minutes). (EDIT) It's interesting that the conclusion $m(E)\leq\lim\inf_{n\to\infty}m(E_{n}$ is basically the set analog of Fatou's lemma. To establish this fact did not require the hypothesis of $E_{n}\subset F$ and $m(F)<\infty$, since it depended on only the upward convergence theorem (akin to the monotone convergence theorem for integrals, which applies only to emph increasing sequences of functions). Establishing $m(E)\geq\lim\sup_{n\to\infty}m(E_{n})$ did, however, and just like for integrals, the hypothesis cannot be dropped. – Taylor Martin Oct 23 '12 at 10:13

Following the hint from Exercise 24 here consider $$F_N := \bigcup_{n \geq N} E_n \Delta E.$$ By monotone convergence for sets $$\lim_N m(F_N) = m(\bigcap_N F_N) = 0,$$ the last equality coming from $\chi_{E_N} \rightarrow \chi_E$ a.e.. Therefore $$0 \leftarrow m(F_N) \geq m(E_N \Delta E) \geq m(E_N \setminus E) = m(E_N) - m(E),$$ thus the conclusion.

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