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Suppose we have function $f_n$ all p-integrable for $p<\infty$. We define new measurable functions $g_n$ as:

$$g_n := \sum_{i=1}^n f_i$$

Further we assume that $g_n\to g$ in $L^p$. Now why is the following true ($\mu# probability measure):

$$ \int{g d\mu} = \int{\lim_{n\to \infty}\sum_{i=1}^n f_id\mu} = \lim_{n\to \infty} \sum_{i=1}^n \int{f_id\mu}$$

The question is, why could we interchange the limit and the integral? If we know that $f_n\ge 0$ this is clear. But what is for example if $f_n$ are random variables?

Thanks for your help.

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In general, a function in $L^p$ may not be integrable. Is the measure space in which the integration is carried out of finite measure? – Julián Aguirre Mar 5 '12 at 16:20
Yes, $\mu$ is a probability measure – user20869 Mar 5 '12 at 16:35
are you assuming that $1\leq p$? – azarel Mar 5 '12 at 17:11
@azarel If $\mu$ is bounded then $L^p\subset L^q$ for all $q<p$. – AD. Mar 5 '12 at 17:15

2 Answers 2

up vote 1 down vote accepted

Call $X$ the measure space, and let $q\ge1$ be such that $1/p+1/q=1$. By Hölder's inequality $$\begin{align*} \Bigl|\int_X(g_n-g)d\mu\Bigr|&\le\int_X|g_n-g|\,d\mu\\ &\le\Bigl(\int_xd\mu\Bigr)^{1/q}\Bigl(\int_X|g_n-g|^pd\mu\Bigr)^{1/p}\\ &=\|g-g_n\|_{L^p(X)}\to0. \end{align*}$$

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Another way to think about this fact: the inclusion $L^p \hookrightarrow L^1$ is continuous (Holder), and $f \mapsto \int f\,d\mu$ is a continuous function from $L^1$ to $\mathbb{R}$. So their composition is also continuous: $g \mapsto \int g\,d\mu$ is a continuous function on $L^p$. Since $g_n \to g$ in $L^p$, we have $\int g_n \to \int g$. – Nate Eldredge Mar 5 '12 at 18:36

By assumption $$\left|\int g-g_n\,d\mu\right|\leq\|g-g_n\|_{L^1}\to0$$ as $n\to\infty$. Now, $$\int g-g_n\,d\mu=\int g\,d\mu - \int \sum_{k=1}^n f_k d\mu=\int g\,d\mu -\sum_{k=1}^n \int f_k d\mu$$

Edit: Above we used a well known property of finite measure spaces -- If $\mu(X)\lt\infty$ and $p>q$ then $f\in L^q$ for all $f\in L^p$, because
$$\int |f|^qd\mu = \left(\int |f|^pd\mu\right)^{q/p}\left(\int 1 d\mu\right)^{1/(p/q)'}=\|f\|_{L^p}^q\,\mu(X)^{1/(p/q)'}$$ by Hölder's inequality.

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