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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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    @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.2012-07-13

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The answer to the question in the title is: No, even on finite measured spaces.

For an example, consider $X=(0,1)$ endowed with the Lebesgue measure, and $f_n=2^n\cdot\mathbf 1_{(0,1/n)}$.

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    Thank you very much. I understand it now. What I asked in the question follows easily from the monotone convergence theorem if $f \in L^p$.2012-07-13