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This are questions

Let $(X,\mathcal{M},m)$ be a measure space. Prove the following by using MCT or DCT:

1) If $\langle A_{n} : n \in \mathbb{N} \rangle$ is an incresing sequence of measurable sets, then $$m \left( \bigcup_{n \in \mathbb{N}} A_{n} \right) = \lim_{n \to \infty} m(A_{n}).$$

2) If $\langle A_{n} : n \in \mathbb{N} \rangle$ is a decreasing sequence of measurable sets with $m(A_{1})< \infty$, then $$m \left( \bigcap_{n \in \mathbb{N}} A_{n} \right) = \lim_{n \to \infty} m(A_{n}).$$

These are my try:

1) Since $\chi_{A_{n}}$ are nonnegative increasing measurable function, thus by MCT, $$\lim_{n \to \infty} m(A_{n}) = \lim_{n \to \infty} \int_{\bigcup_{n \in \mathbb{N}} A_{n}} \chi_{A_{n}} = \int_{\bigcup_{n \in \mathbb{N}} A_{n}} \lim_{n \to \infty}\chi_{A_{n}} = \int_{\bigcup_{n \in \mathbb{N}} A_{n}} \chi_{\bigcup_{n \in \mathbb{N}} A_{n}} = m \left( \bigcup_{n \in \mathbb{N}} A_{n} \right).$$

2) Now, $|\chi_{A_{n}}| \leq 1$ and $\lim_{n \to \infty} \chi_{A_{n}} = \chi_{\bigcap_{n \in \mathbb{N}}A_{n}}$. By DCT, we get that $$\lim_{n \to \infty} m(A_{n}) = \lim_{n \to \infty} \int_{A_{1}} \chi_{A_{n}} = \int_{A_{1}} \lim_{n \to \infty}\chi_{A_{n}} = \int_{A_{1}} \chi_{\bigcap_{n \in \mathbb{N}} A_{n}} = m \left( \bigcap_{n \in \mathbb{N}} A_{n} \right).$$

Is it correct in 1)?

I'm not sure in 2) because the condition $m(A_{1}) < \infty$ does not use.

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    2) can't be proven without using $m(A_1)<\infty.$ For example on the reals let $A_n=(n,\infty).$ – 2017-01-12
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    It also needs the space to satisfy the DCT. For example $X=\mathbb R$ and $M$ is generated by $\{(0,r):r>0\},$ and $m(U)=1$ if $\exists r>0\; (U\supset (0,r),)$ otherwise $m(U)=0.$ This measure space has atoms and does not satisfy DCT. E,g, $A_n=(0,1/n).$ – 2017-01-12

0 Answers 0