Let $(X,\Sigma, \mu)$ be a measurable space and let $f: X \mapsto \mathbb{R}$ be an integrable function. I want to show that for $\epsilon > 0$, I can find a simple function g such that $\int_{X} |f - g| d\mu < \epsilon$.

Since f is integrable, we know that $\int f = \text{sup}\{\int g :\text{g is a simple function and } g \leq_{a.e} f\}$ so if $g \leq_{a.e} f$ we have the inequality $\int f d\mu > \int g d\mu = \int \sum_\limits{i=1}^{n} a_{i} \chi(E_{i})d\mu$ where each $E_{i}$ are measurable sets.

But now I'm not sure how to proceed.

  • $\begingroup$ You have only used the fact that sup is an upper bound on $\{\int g: g \ {\rm simple \ and} \, g \leq f \}$. Now use the fact that sup is the least upper bound. $\endgroup$ – Kenny Wong Mar 20 '17 at 11:45
  • $\begingroup$ So if we have another function, say $h$, such that $g \leq h$ then we have that $\int g \leq \int f \leq \int h$? $\endgroup$ – Arsalan Dashti Mar 20 '17 at 11:58
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    $\begingroup$ Your definition of $\int f$ only works for nonnegative $f$ and needs another clause: $0\leq g\leq f$. $\endgroup$ – drhab Mar 20 '17 at 12:00
  • $\begingroup$ I'm not really sure how to bring $\epsilon$ into any of this $\endgroup$ – Arsalan Dashti Mar 20 '17 at 12:08

Start by writing $f=f_+-f_-$ where $f_+(x):=\max(f(x),0)$ and $f_-(x):=\max(-f(x),0)$.

Then $f_+,f_-$ are nonnegative functions that inherit measurability and integrability from $f$.

By definition: $$\int f_+=\sup\left\{\int s\mid s\text{ is a simple function with }0\leq s(x)\leq f_+(x)\text{ for every }x\in X\right\}$$

So for any $\epsilon>0$ we can find a simple function $g_1$ with $0\leq g_1(x)\leq f_+(x)$ for every $x\in X$, and: $$\int g_1\leq\int f_+<\frac12\epsilon+\int g_1$$

Likewise we can find a simple function $g_2$ with $0\leq g_2(x)\leq f_-(x)$ for every $x\in X$, and: $$\int g_2\leq\int f_-<\frac12\epsilon+\int g_2$$ Then $g:=g_1-g_2$ is a simple function with $(f-g)_+=f_+-g_1$ and $(f-g)_-=f_--g_2$.

Then: $$\int|f-g|=\int (f-g)_++\int(f-g)_-=\int (f_+-g_1)+\int (f_--g_2)<\frac12\epsilon+\frac12\epsilon=\epsilon$$

  • $\begingroup$ @KennyWong Yes, thank you. I will repair. $\endgroup$ – drhab Mar 20 '17 at 12:51
  • $\begingroup$ Wow, thank you. A very simple and elegant solution. $\endgroup$ – Arsalan Dashti Mar 20 '17 at 13:19

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