Let $(X, B, \mu)$ be a measure space. $f$ be an integrable function on $X$ and let
$$
G(E) := \int_E f(x) d \mu(x)
$$
for $E \in B.$
I want to show that for measurable function $g$,
$$
\int_X g(x) d G(x) = \int_X g(x)f(x) d\mu (x)
$$
holds. Actually I cannot understand what does the notation $\int.. dG(x)$ means. Please give me some help.
$\int_X g(x) d G(E) = \int_X g(x)f(x) d\mu (x)$?
0
$\begingroup$
real-analysis
integration
measure-theory
lebesgue-integral
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1$G$ is a measure. That's $\int\cdots dG(x)$. It means integration with respect to the measure $G$. – 2017-01-08
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0@OpenBall oh thanks, but how can I show that the equality holds? – 2017-01-08
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0@M.Doe From openballs comment, can you see that G \|| \mu? – 2017-01-08