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Let $(X,A,\mu)$ be a measure space and let $(f_n)_{n\in\mathbb N}$ a sequence of positive measurable functions s.t. $$ f_1(x)\ge f_2(x) \ge \ldots \ge f_n(x)\ge ... \ge 0, \qquad \forall x\in X $$ and $f_n(x) \to f(x)$ for every $x \in X$. If $f_1 \in L^1(X)$ then $$ \lim_n \int_X f_n d\mu = \int_X f d\mu. $$

I think this is quite immediate from dominated convergence theorem. Indeed, every $f_n$ is dominated, i.e. $\vert f_n(x) \vert =f_n(x)\le f_1(x)$, for every $x \in X$. Since $f_1 \in L^1$, we have $\int_X f_n d\mu \to \int_X fd\mu$.

Is this correct?

Indeed, I have another proof of this fact: let's define $g_n(x)=f_1(x)-f_n(x)\ge 0$, $g_n(x)\le g_{n+1}(x)$ and $g_n(x)\to f_1(x)-f(x)$, for every $x \in X$. Then by monotone convergence, we get $$ \lim_n \int_X g_n(x) d\mu = \int_X f_1(x)-f(x)d\mu $$ hence (since $f_1 \in L^1$, i.e. $\int_X f_1 d\mu < \infty$) we get - by subtraction - $$ \lim_n \int_X f_n(x) d\mu = \int_X f(x)d\mu $$

Are both proofs correct? Thanks in advance.

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I don't see where $f_1$ is assumed integrable (and we need this, otherwise in the real line with Lebesgue measure, taking $f_n=\chi_{(n,+\infty)}$). –  Davide Giraudo Nov 10 '12 at 17:39
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Sorry Davide, there was a typo in the text of the problem. Look at it now. I apologize for my mistake. –  Romeo Nov 10 '12 at 17:51
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1 Answer

up vote 1 down vote accepted

Both proofs are correct. I personally prefer the second one, as it involves only monotone convergence, hence it's a little less advanced.

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Thanks a lot for your help. –  Romeo Nov 10 '12 at 18:07
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