The problem:
Suppose $E \in \mathfrak{M}$ and that $f$ is (Lebesgue) integrable over $E$. For any $\epsilon > 0$ show that there exist simple, step, and continuous functions $\varphi, \psi, g$ (respectively) such that:
- $\displaystyle\int_E |f – \varphi|\,dm < \epsilon$
- $\displaystyle\int_E|f – \psi|\,dm < \epsilon$
- $\displaystyle\int_E|f – g|\,dm <\epsilon$
Now, we have shown in the notes that for $f: [a, b] \to [-\infty, +\infty]$ measurable (and $f \neq \pm \infty$ almost everywhere) we can find simple, step, and continuous functions such that :
$$m(\left \{x \in [a, b] : |f(x) – (\text{the function})| \geq \epsilon \right \}) < \epsilon $$
So I'm certain that these two are related!
We also have that $\displaystyle\int_E |f|\,dm < \infty$ by assumption on $f$.
But how are these connected?
Best Answer
Here's a condensate of some aspects of our long discussion elsewhere.
Let us do the approximation by simple functions first:
Assume $f \geq 0$. From the definition of the Lebesgue integral we have $$ \int f = \sup{\left\{ \int s \,:\, s \leq f,\;s\text{ is simple}\right\}}. $$ If $f$ is integrable, then $\int f \lt \infty$. For every $\varepsilon \gt 0$ we can find a simple function $0 \leq s \leq f$ such that $$ \int f \lt \varepsilon + \int s, $$ by definition of the supremum, so $$ 0\leq \int (f-s) \lt \varepsilon. $$
If $f$ is integrable, but not necessarily non-negative, split it into positive and negative parts $f = f_+ - f_-$, where $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$. Given $\varepsilon \gt 0$ choose simple functions $0 \leq s_{\pm} \leq f_{\pm}$ such that $$ 0 \leq \int (f_{\pm} - s_{\pm}) \lt \varepsilon/2 $$ then put $s = s_{+} - s_{-}$ and observe $$ 0 \leq \int |f - s| = \int |f_+ - f_- + s_+ - s_-| \leq \int |f_+ - s_+| + \int |f_- - s_-| \lt \varepsilon. $$
In other words, your first bullet point is immediate from the definitions of the integral and the word “integrable”. Note also that this works on any measure space, we haven't used any special property of Lebesgue measure.
I find it easier and more instructive to do the case of continuous functions before dealing with step functions.
We are going to exploit the following basic property of Lebesgue measure $m$:
Given this, it is easy to write down a continuous function $\varphi: \mathbb{R} \to [0,1]$ such that $\varphi|_{F} = 1$ and $\varphi|_{U^c} = 0$ while $0 \lt \varphi(x) \lt 1$ for $x \notin F \cup U^c$—a simple version of Urysohn's lemma for metric spaces:
Indeed, recall that in a metric space the distance function $x \mapsto \mathrm{dist}(x,A) = \inf_{a \in A} d(x,a)$ of $x \in X$ to a closed set $A$ is continuous (even $1$-Lipschitz) and put $$ \varphi(x) = \frac{\mathrm{dist}(x,U^c)}{\mathrm{dist}(x,F)+\mathrm{dist}(x,U^c)}. $$ For the characteristic function $[E]$ of $E$ we then have $$ 0 \leq \int |[E]-\varphi| \leq m(U \smallsetminus F) \lt \varepsilon. $$ This implies that for every (integrable) simple function $s$ and every $\varepsilon \gt 0$ there is a continuous function $\varphi$ such that $\int |s - \varphi| \lt \varepsilon$. Finally, for a general integrable function $f$ we can find a simple function $s$ such that $\int |f - s| \lt \varepsilon /2$ by the first part of this answer, and then we can find a continuous function $\varphi$ such that $\int |s-\varphi| \lt \varepsilon /2$, hence:
Finally, to get the result for step functions, recall from Riemann integration theory that a continuous function (on a bounded set) is a uniform limit of step functions.