# Proof verification that $\mu_f$($E$) $=$ $\int_E f\,d\lambda$ is a measure on $\mathcal M$.

I was wondering if I had correctly proved this problem: Let $f$ be a nonnegative $\mathcal M$-measurable function. Define $\mu_f$ on $\mathcal M$ by $\mu_f$($E$) $=$ $\int_E f\,d\lambda$. Prove that $\mu_f$ is a measure on $\mathcal M$.

Here is my solution:

Condition (1): Since $f$ is nonnegative, $f$ $\ge$ $0$. Also since $f$ is $\mathcal M$-measurable function, we get for each $E$ $\in$ $\mathcal M$ that $\int_E f\,d\lambda$$\ge 0. Hence, \mu_f(E) \ge 0 for each E \in \mathcal M. Condition(2): Let E = \emptyset. Since \lambda(E) = 0, we have that \int_E f\,d\lambda = 0. So, \mu_f(\emptyset) = 0. Condition (3): Let \{E_n\}_{n=1}^{\infty} be a sequence of pairwise disjoint sets in \mathcal M. Now because we also have that E_n \in \mathcal M for each n \in \Bbb N, and f is a nonnegative \mathcal M-measurable function, we have that \int_{\bigcup_{n=1}^{\infty}E_n}f{\rm d}\lambda = \sum_{n=1}^{\infty}\int_{E_n}f. Hence, \mu_f({\bigcup_{n=1}^{\infty}E_n}) = \sum_{n=1}^{\infty}(\mu_f(E_n)) ## 1 Answer The only issue is condition (3), as you are doing nothing. Unless you already have a previous proof of$$\int_{\bigcup_{n=1}^{\infty}E_n}f\,{\rm d}\lambda=\sum_{n=1}^{\infty}\int_{E_n}f.$$The justification is not hard, using monotone convergence, but it needs to be done. You have$$\tag1\int_{\bigcup_{n=1}^{\infty}E_n}f\,{\rm d}\lambda =\int f\,1_{\bigcup_nE_n} =\int\sum_{n=1}^\infty f\,1_{E_n} =\sum_{n=1}^{\infty}\int_{E_n}f.$$The nontrivial step is the last one, where one uses monotone convergence. • Is that really necessary? The sets$E_n$are pairwise disjoint, so integral over their union must be the sum of integral over$E_n$(this follows from definition of integration) – Prasun Biswas Mar 11 '18 at 4:27 • Or is it because the sequence$\{E_n\}$is not finite that we need to do some justification? – Prasun Biswas Mar 11 '18 at 4:29 • The textbook I'm using gave a proof of$\int_{\bigcup_{n=1}^{\infty}E_n}f\,{\rm d}\lambda=\sum_{n=1}^{\infty}\int_{E_n}f$. The reasoning I have in (3) is simply listing the conditions needed to use that fact. – FoxViking Mar 11 '18 at 5:16 • @PrasunBiswas: I don't think so. The definition of integral gives you the first equality in$(1)\$ above; you still need to do the rest. – Martin Argerami Mar 11 '18 at 5:22
• @PrasunBiswas The thing that follows from the definition of integration is that the integral over the finite disjoint union of measurable sets is the sum of integrals. The case where we take infinitely many disjoint sets requires an application of the monotone convergence theorem. – Alex Ortiz Mar 14 '18 at 0:35