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I was working on the folllowing exercise from Tao's Measure Theory book:

We say that a sequence $E_n$ of sets in $R^d$ converges pointwise to another set $E$ if the indicator functions $1_{E_n}$ converge pointwise to $1_E$. Show that if $E_n$ are all Lebesgue measurable and converge pointwise to E then E is measurable also. He gives the hint to use the identity $1_{E(x)} =$ lim $ inf_{n - > \infty} 1_{E_n(x)} $to write E in terms of countable unions and intersections of $E_n$.

If instead we changed the assumptions such that the indicator functions converged uniformly to $1_E$ on the domain $\cup E_n \bigcup E$ then there exists an $n$ such that $E = \cup_{m > n} E_m$. However I am not quite sure how to work with the weaker assumption of pointwise convergence. Any help would be appreciated!

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This is basically a baby version of the following result (you can skip to bottom).

If $(f_n)$ are measurable and $f_n \to f$ pointwise, then $f$ is measurable.

Why ? Because of pointwise convergence $$\{x:f(x)> a\}=\cup_{n=1}^{\infty} \{x: f_m(x) > \alpha \mbox{ for all } m \ge n\}.$$

Read the RHS as all $x$ such that $f_m(x)>\alpha$ for all $m$ (possibly depending on $x$) large enough.

We can rewrite the RHS as

$$ \cup_{n=1}^\infty \cap_{m\ge n} \{x:f_m(x)>\alpha\}.$$

The sets on the RHS are all measurable. The result follows. Note that enough to consider $\alpha=0$, in which case the set $\{x:f_m(x)>\alpha\}$ is $E_m$. In other words,

$$E = \cup_{n=1}^\infty \cap_{m\ge n} E_m,$$

and this identity is all you need, and which you can derive directly (with the same argument -- just follows from the definition of pointwise convergence: a point is in $E$ if and only if for all $m$ large enough belongs to $E_m$).

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  • $\begingroup$ Maybe you could explain why an intersection pops up starting from the 2nd equation. $\endgroup$ – zebullon Mar 17 '16 at 3:14
  • $\begingroup$ "For all $m$" means intersection. $\endgroup$ – Fnacool Mar 17 '16 at 3:15

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