# Integral equality including function with compact support

I don't understand a proof step in my measure theory book, maybe you can help me. We have a Radon (outer) measure $\mu: \mathcal P(\Bbb R^n) \to [0, \infty]$ and a $f \in L^1_{\text{loc}}(\Bbb R^n)$, such that $$\mu(A) = \int_A f \, \mathrm dx \quad \text{for each Borel set } A \subset \Bbb R^n$$

Now, the proof states, that if $\phi \in C_c(\Bbb R^n)$, then $$\int_{\Bbb R^n} \phi f \, \mathrm dx = \int_{\Bbb R^n} \phi \, \mathrm d\mu \; ,$$ where the first integral is taken over the normal Lebesgue measure. Can somebody explain me, why this holds?

That's approximation: Let $\phi \ge 0$ (The general case is done by writing $\phi = \phi^+ - \phi^-$). Let $s_n$ be a sequence of nonnegative simple functions so that $s_n \to \phi$ almost everywhere from below. As $|s_n f| \le |\phi f|$ and $\phi f$ is integrable, the Lebesgue's dominated convergence theorem states

$$\int_{\mathbb R^n} \phi f dx = \lim_{n\to \infty} \int s_n fdx.$$

Write $s_n = \sum_k a_{n,k} \chi_{A_{n,k}}$, then

$$\int s_n fdx = \sum_k a_{n,k}\int_{A_{n,k}} fdx = \sum_k a_{n,k} \mu(A_{n,k})\to \int_{\mathbb R^n} \phi d\mu.$$

• Thanks for this, it's clear now. I think in the first line you mean $\phi \geq 0$. – aexl Nov 5 '15 at 13:40
• Yes, @aexl. Edited. – user99914 Nov 5 '15 at 13:41