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Is this a valid proof of the Monotone convergence theorem for Lebesgue measurable sets?

Let $E_1 \subseteq E_2 \subseteq \dots $ be a sequence of Lebesgue measurable sets $E_i$. Then we want to show that $\underset{N \rightarrow \infty } \lim m(E_N) = m(\cup_{i=1}^\infty E_i)$.

Proof: Assuming each $E_i$ has finite measure, then $m(\cup_{i=1}^\infty E_i) = m(E_1) + (m(E_2) - m(E_1)) + (m(E_3) - m(E_2)) \dots = \underset{N \rightarrow \infty } \lim \sum_{i=1}^N m(E_i) - \underset{N \rightarrow \infty } \lim \sum_{i=1}^{N-1} m(E_i) = \underset{N \rightarrow \infty} \lim m(E_N)$

And then we treat the case where one of the $E_i$'s has infinite measure separately. Having forgotten basic calculus and convergence of sequences and similar, I am mostly wondering whether these steps are valid.

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It has some problems.

(1) Never mind. Nate Eldredge has set me straight on this.

(2) $\lim_{N\to\infty} \sum_{i=1}^N m(E_i) - \lim_{N\to\infty} \sum_{i=1}^{N-1} m(E_i)$ is quite evidently $0$ (since both limits converge to the same value). Of course, your problem here is just that you should not have broken the limit over the subtraction. You want just $$\lim_{N\to\infty}\left( \sum_{i=1}^N m(E_i) - \sum_{i=1}^{N-1} m(E_i)\right)$$

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  • $\begingroup$ For (1), the statement $m(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty m(A_i)$ is simply the countable additivity of the Lebesgue measure $m$. This is usually proved from first principles as part of the construction of Lebesgue measure, and is needed in proving the monotone convergence theorem for measurable functions. $\endgroup$ – Nate Eldredge Dec 2 '15 at 18:27
  • $\begingroup$ @NateEldredge - Thank you. I should have remembered that. And for measures in general, countable additivity is one of the axioms of measure theory. $\endgroup$ – Paul Sinclair Dec 2 '15 at 18:51

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