I'm reading a proof in a book and don't understand a crucial step.

So we have $f \in L^1_{\text{loc}}(\mathbb R^n)$ and $\phi \in C_c(\mathbb R^n)$. Now $\mu$ is a Radon measure, such that $$ \mu(A) = \int_A f \; \mathrm dx $$ for every Borel set $A \subset \mathbb R^n$.

Now the proof states that $$ \int_{\mathbb R^n} \phi f \; \mathrm dx = \int_{\mathbb R^n} \phi \; \mathrm d\mu \; .$$ Why is this true? Is the author using a Theorem that I should know? Or is it obvious from the upper equation? I don't see it...


That's a standard argument in measure theory: First by taking $\phi = \phi^+ - \phi^-$, we assume that $\phi$ is nonnegative. By definition, your assertion is true when $\phi = \chi_A$ for any Borel set $A$. Thus it is also true when $\phi$ is a simple function.

In general if $\phi$ is continuous with compact support, let $\phi_n$ be a sequence of nonnegative simple functions so that $\phi_n$ increases to $\phi$ a.e.. Then also $\phi_n f$ increases to $\phi f$ a.e. (assume that $f\ge 0$). By using monotone convergence theorem twice, we have

$$ \int \phi f dx = \int \lim (\phi_n f) dx = \lim \int \phi_n fdx = \lim \int \phi_n d\mu = \int \lim \phi_n d\mu = \int \phi d\mu.$$

(For general $f$, again consider $f = f^+ - f^-$).

  • $\begingroup$ Thank you for this, great answer! $\endgroup$ – aexl Jan 18 '15 at 13:49

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