Let $(X,\mathcal{M},\mu)$ be a measure space, and let $f:X\to [0,\infty]$ be a measurable function. Define the map $\lambda:\mathcal{M}\to[0,\infty]$, $\lambda(E)=\int_X \chi_E\ f d\mu$, where $\chi_E$ denotes the characteristic function of $E$.
Show that for any measurable function $g:X\to[0,\infty]$, one has $\int_X gd\lambda=\int_X gf d\mu$ in $[0,\infty]$.
Is there any neat way to prove the above? I tried using the definition (sup of simple functions) and it got quite messy.
Thanks for any help.
(We may use the fact that $\lambda$ is a measure, and it is absolutely continuous with respect to $\mu$).
Answer
If $g\geq 0$, there is a sequence $(s_n)$ of non-negative simple functions such that $s_n\nearrow g$. Since $\lambda$ is a positive measure,
$$
\lim_{n\to \infty} \int_X s_n d\lambda = \int_X gd\lambda
$$
and since $s_nf \nearrow gf$, it follows that
$$
\lim_{n\to\infty}\int_X s_nf d\mu = \int_X gfd\mu
$$
Now use the fact the corresponding terms in the limit are equal.
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