# $L^p$ convergences implies to interchange integral and sum

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