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I was reading through Royden, Bartle and other Measure Theory textbooks and I saw this problem more than once. If $f \in L(X,\mathcal{A},\mu)$ and $\varepsilon > 0$, then there exists a simple function $\varphi$ such that

$$\int |f-\varphi| \; d\mu < \varepsilon.$$

I keep having trouble setting up an inequality that can satisfy this so any help is appreciated.

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Hint: How is the integral of an arbitrary measurable function defined? –  Chris Janjigian Apr 11 '12 at 15:20
    
Look at the definition of Lebesgue integral, and also the Dominated Convergence Theorem. –  Martin Argerami Apr 11 '12 at 15:31

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up vote 3 down vote accepted

First you show that for a non-negative measurable function $f: X \to [0, \infty]$ there exists a sequence of non-negative simple functions $(s_n)$ such that $s_n \leq s_{n+1}$ and $\lim_{n \to \infty} s_n(x) = f(x)$ pointwise for all $x \in X$.

To show this you can construct $s_n$ explicitly as follows (construction taken from here):

$$ s_n(x) = \begin{cases} n & \text{ if } f(x) \geq n \\  \frac{i-1}{2^n} & \text{ if } \frac{i-1}{2^n} \leq f(x) \leq \frac{i}{2^n} \text{ where } 1 \leq i \leq n2^n \end{cases}$$

Then apply the monotone convergence theorem to get $\int_X f d \mu = \lim_{n \to \infty} \int_X s_n d \mu $.

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