Countable additivity of Lebesgue integral question

646 Views Asked by At

I want to understand whether the following holds for all integrable $\varphi\in \mathcal L^1(\Omega, \mathcal A, P):$

Let $A = \bigcup_{n=1}^\infty A_n$ be a disjoint union of measurable sets $A_n \in \mathcal A$. Then $$\int_A \varphi \,dP = \sum_{n\geq 1}\int_{A_n}\varphi \, dP.$$

Now the Proof I saw writes $$1_{A} = 1_{\bigcup_{n=1}^\infty A_n} = \sum_{n=1}^\infty 1_{A_n}$$ since the $A_n$ are disjoint. Then the interchange of integral and limit is justified by the monotone convergence theorem. But I think that this is not applicable for negative $\varphi$ since the partial sums then must not be increasing. Is the statement wrong in this form or is there an alternative way of proving it?

1

There are 1 best solutions below

3
On BEST ANSWER

This is justified by the dominated convergence theorem, not the monotone convergence theorem. Just apply it to the partial sums $\sum_{n=1}^N1_{A_n}\varphi$, which are bounded in absolute value by the integrable function $|\varphi|$.