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1) Suppose that $A_n$ is measurable with $\mu(A_n) < \infty$ for all $n$ and $1_{A_n}$ converges in measure to $f$, then there exists a measurable set $A$ such that $f = 1_A$

My idea in doing this problem is :

Since $1_{A_n}$ converges in measure to $f$, there exist a subsequence $1_{A_{n_j}}$ converging to $f$ a.e. I think that I should set $A = \cap_{j \in \mathbb{N}} A_{n_j}.$ It is clear that $A$ is measurable. Anyway, I have 2 concerns :

  1. Is $A$ empty ? Should I care if $A$ is empty ? (Empty is a measurable set, but it is strange to write $f = 1_\phi$)

  2. It seem to me that $1_A = f$ a.e. But I find that I am not sure how to prove this rigorously (I have a sense that I might need Dominated convergence thm for this)

2) Change of variable : If $f$ is an integrable function on reals and $a$ is non-zero real number, then $$\int_{\mathbb{R}}f(x+a)\ dx = \int_{\mathbb{R}}f \ dx$$ $$\int_{\mathbb{R}}f(ax) \ dx = |a^{-1}|\int_{\mathbb{R}}f \ dx.$$

I do not sure how to begin. Since $f$ is reals, I should first prove the result for non-negative function. Then it seems that I need to establish this result for an indicator function, then simple function. But it looks complicated. Could anyone give a hint or outline the proof ?

Thank you very much

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    $\begingroup$ As for (1), the limit of a sequence of zeros and ones can only be $0$ or $1$, so just define $A$ as the set where the pointwise limit is one. $\endgroup$ – user138530 Oct 13 '15 at 4:40
  • $\begingroup$ (1) Notice that your guess about the limit set $A$ is not correct: consider the intervals $A_n = \left(0, 1-\frac{1, n+1}\right)$: the corrisponding indicator functions converge (in $L^1$) to $1_{(0, 1)}$, that is not the set you expected. Concern 1: no problem, the indicator function of the empty set is the null function. $\endgroup$ – SiD Oct 13 '15 at 11:49
  • $\begingroup$ @SiD According to your example, it seems that I might try A equals union of $A_{n_j}$ instead. But if I change your example to $A_n = (1, 1 + \frac{1}{n+1})$, then it seems that A is the intersection, not union. So I wonder how I should define A ? The set where pointwise limit is 1 ? Then why A is measurable ? I am fo not quite comprehend the intuition of defining A to be this set. Can you explain more ? $\endgroup$ – Both Htob Oct 14 '15 at 10:10

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