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Going through a proof in the measure theory book by Donald Cohn, me and my mate are disagreeing about the validity of a step in my argument...

The relevant assumption to my argument is

Let $\{A_n\}$ be sets in an arbitrary measurable space such that $\lim 1_{A_n}(x) \rightarrow 0 $ $\mu$-almost everywhere.

I then define $g(x):=\lim 1_{A_n}(x)$ and $h(x):=0$ then $g(x)=h(x)$ $\mu$-almost everywhere (that is pretty much just rewriting the assumptions)

$\int h(x) d\mu =0$ implies that the integral of $g(x)$ exists and $\int h(x) d\mu=\int g(x) d\mu =0$ a.e, translating back to the original notation $\int \lim 1_{A_n}(x) d\mu=0$ a.e

My classmate tells me that this argument is wrong and that I can not use the Proposition 2.3.9 in the book (which states basically that for measurable functions agreeing a.e, if the integral of one exist, then both exist and are equal a.e)

Who is right ? Thanks.

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    Why do you write that two integrals are equal a.e.? How is $g$ defined on the set $\{\lim_n 1_{A_n} \not \to 0 \}$? What exactly is your classmate's objection? Why does s/he think the Proposition can't be applied? – 2017-01-21

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