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Let $f: X \to [0, \infty) \subset \mathbb R$ measurable where $X$ is a measure space. Let $f_n : X \to [0, \infty) $ be simple functions (i.e. linear combinations of characteristic functions of measurable sets) such that for each $x \in X$, $f_n(x) \leq f_{n+1}(x)$ and $f_n(x)$ converges to $f(x)$.

How can I prove that $$ \|f_n - f \|_p = \left ( \int_X |f - f_n|^p d \mu\right )^{1/p} \xrightarrow{n \to \infty} 0$$

I don't think this is right but if for $n > N_x$, $|f_n(x) - f(x)| \leq \varepsilon$, we can let $N = \sup_{x \in X} N_x$ to get $\|f_n - f\|_\infty \leq \varepsilon$ and then $$ \|f_n - f \|_p = \left ( \int_X \|f - f_n\|^p d \mu\right )^{1/p} \leq \left ( \int_X \varepsilon^p d \mu\right )^{1/p} = \mu(X)^{1/p} \varepsilon $$

But $\mu(X)$ could be infinite so I'm not sure what to do. Thanks.

Edit What assumptions do I need to make this true?

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Since $x \mapsto e^x$ is measurable, there exists a sequence of simple functions that converges to this exponential on $[0,+\infty)$. Of course the exponential is not integrable. Are you forgetting an additional assumption? – Siminore Jul 13 '12 at 14:58
@Siminore Yes, $f_n \leq f_{n +1}$. – Rudy the Reindeer Jul 13 '12 at 15:01
@bananalyst: Please edit the question to add this statement. – Nate Eldredge Jul 13 '12 at 15:17
@bananalyst: Something is still missing; Siminore's counterexample still applies. You can find simple functions in $L^p$ that increase to $e^x$. So even with the additional statement this is not true. – Nate Eldredge Jul 13 '12 at 15:53
up vote 2 down vote accepted

The answer to the question in the title is: No, even on finite measured spaces. For an example, consider $X=(0,1)$ with Lebesgue measure and $f_n=2^n\cdot\mathbf 1_{(0,1/n)}$.

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I forgot to write it in the question but I meant $f_n(x) \leq f_{n + 1}(x)$ for all $x \in X$. Should I post another question? – Rudy the Reindeer Jul 13 '12 at 15:01
Then if $f_1$ is in $L^p$, this follows from Lebesgue dominated convergence theorem. – Did Jul 13 '12 at 15:05
Did you mean if $f_n$ is in $L^p$? – Rudy the Reindeer Jul 13 '12 at 15:08
No. $ $ $ $ $ $ – Did Jul 13 '12 at 15:10
Right. I think I see it now. We can swap limit and integral and get the integral over the zero function. – Rudy the Reindeer Jul 13 '12 at 15:10

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