Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of measures on $A$ satisfying $\mu_n \nearrow \mu$ and $f,f_n \in M^+$ ( $n \in \mathbb{N}$ ) satisfying $f_n \nearrow f$. Show that $$ \lim_{n \rightarrow \infty} \int_{\Omega} f_n\, d\mu_n = \int_{\Omega} f\, d\mu. $$

So my idea is to show this claim for a fixed $f_n$. Maybe we could use this then for the claim itself. Moreover I think that monotone convergence theorem could be helpful.

Remark: $M^+ $ := {$ f | f$ is measurable and non-negative }


So $\mu_{n}$ is absolutely continuous with respect to $\mu$, so $\dfrac{d\mu_{n}}{d\mu}$ exists, and we see that \begin{align*} \mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu\leq\mu_{n+1}(A)=\int_{A}\dfrac{d\mu_{n+1}}{d\mu}d\mu, \end{align*} so \begin{align*} \int_{A}\left(\dfrac{d\mu_{n+1}}{d\mu}-\dfrac{d\mu_{n}}{d\mu}\right)d\mu\geq 0 \end{align*} holds for every $A$, so \begin{align*} \dfrac{d\mu_{n+1}}{d\mu}\geq\dfrac{d\mu_{n}}{d\mu},~~~~\text{a.e.}. \end{align*} Since \begin{align*} \mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} we take $n\rightarrow\infty$ and using Monotone Convergence Theorem to deduce that \begin{align*} \mu(A)=\int_{A}\lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} this shows that \begin{align*} \lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}=1,~~~~\text{a.e}. \end{align*} Now we write that \begin{align*} \int_{\Omega}f_{n}d\mu_{n}=\int_{\Omega}f_{n}\dfrac{d\mu_{n}}{d\mu}d\mu, \end{align*} and use the fact that $f_{n}\dfrac{d\mu_{n}}{d\mu}\uparrow f$ a.e. to get the result.

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  • $\begingroup$ Thank you for your answer user284331, but I have two questions: what do you mean with $ \frac{d\mu_n}{d\mu} $? and why is $\mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu $ ? I think the rest is clear then. $\endgroup$ – RukiaKuchiki Apr 25 '18 at 18:07
  • $\begingroup$ This is our definition of absolutely continuous: Let µ and ν be measures on a measurable space . We call $ν$ absolutely continuous with respect to $µ$ if for all A ∈ $A$, $µ$[A] = 0 ⇒ $ν$[A] = 0 . $\endgroup$ – RukiaKuchiki Apr 25 '18 at 18:10
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    $\begingroup$ There is a theorem called Radon-Nikodym which saying that formula. $\endgroup$ – user284331 Apr 25 '18 at 18:12

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