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This is question 5.D in Bartle's Elements of Integration.

If $f \in L(X,\mathcal X,\mu)$ and $\epsilon > 0$, then there exists a $\mathcal X$ − measurable simple function $\phi$ such that: $\int|f-\phi|d\mu<\epsilon$.

The answer seems trivial when $f$ is either (a.e.) positive or (a.e.) negative. However, when it is both positive and negative, I cannot define simple function that works for both cases.

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    $\begingroup$ Hint: $f= \max(f,0)-\max (-f,0)$. . $\endgroup$ – Fnacool Apr 27 '16 at 2:44
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    $\begingroup$ I tried that. Approximate the $f^+$ and $f^-$ through non-negative simple functions $\phi^+$ and $\phi^-$, but this forces me to define $\phi =\phi^+-\phi^- $ which is not simple because it is not positive. $\endgroup$ – julian.marr Apr 27 '16 at 2:47
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    $\begingroup$ en.wikipedia.org/wiki/Simple_function $\endgroup$ – Fnacool Apr 27 '16 at 2:51
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    $\begingroup$ So, $\phi$ is a simple function since it takes only a finite number of values. Thank you. Note myself: read definitions carefully in the future. $\endgroup$ – julian.marr Apr 27 '16 at 2:58
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Begin by noticing that $f=f^+-f^-$ where$$ f^+(x)=\max\{f(x),0\}\text{ and }f^-(x)=\max\{-f(x),0\} $$and, similarly, $\varphi=\varphi^+-\varphi^-$ for a simple function $\varphi$. Since the function $f^+,f^-,\varphi^+$ and $\varphi^-$ are all nonnegative, the following results are trivial$$ \int |f^+-\varphi^+|\,d\mu<\frac{\varepsilon}{2}\text{ and }\int |f^--\varphi^-|\,d\mu<\frac{\varepsilon}{2} $$for any $\varepsilon>0$. Then, we have$$ \int |f-\varphi|\,d\mu\leq \int |(f^+-f^-)-(\varphi^+-\varphi^-)|\,d\mu\leq \int |f^+-\varphi^+|\,d\mu+\int |f^--\varphi^-|\,d\mu<\varepsilon. $$

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  • $\begingroup$ Why $\int |f^+ - \phi^+| d\mu < \frac{\varepsilon}{2}$? I know that given $\varepsilon > 0$, exist a simple function $\phi^+$ such that $|f^+ - \phi^+| < \frac{\varepsilon}{2}$, then $\int |f^+ - \phi^+| d\mu < \frac{\varepsilon}{2} \int d\mu \Longrightarrow \int |f^+ - \phi^+| d\mu < \frac{\varepsilon}{2} \mu(X)$. I think it's not possible choose $\varepsilon = \frac{\alpha}{\mu(X)}$ for an arbitrary $\alpha > 0$ unless $\mu(X) < + \infty$, so why $\int |f^+ - \phi^+| d\mu < \frac{\varepsilon}{2}$? $\endgroup$ – George Jan 18 '18 at 21:43

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