Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
The Chaz: what is $\mathfrak{M}$ and what is $\bullet m$? Lebesgue measurable sets and Lebesgue (outer?) measure? –  t.b. Oct 29 '11 at 15:56
1  
There's probably a better format for bullet points, but by the time I had typed that much, my laptop was already getting bogged-down (only happens when compiling on this site-???)! –  The Chaz 2.0 Oct 29 '11 at 17:36
2  
I know exactly what you're talking about. Has to do with their desire to render the code instantly. You may want to try Jack's bookmarklets for improving performance. –  t.b. Oct 29 '11 at 17:43
2  
What definition of the Lebesgue integral did you learn? I know the definition of the Lebesgue integral as the sup of the integral on all simple function limiting the function from below - then you get the simple function "free". And isn't a simple function a particular case of a step function? –  yaakov Oct 29 '11 at 19:11
1  
@yaakov: we have that definition, as well as two other (equivalent, surely) definitions. A step function is a special case of a simple function methinks. –  The Chaz 2.0 Oct 29 '11 at 19:51

2 Answers 2

up vote 8 down vote accepted

Here's a condensate of some aspects of our long discussion elsewhere.

Let us do the approximation by simple functions first:

  1. 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 $$ 0\leq \int (f-s) \lt \varepsilon. $$

  2. 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 $m$:

Regularity of Lebesgue measure: For every measurable set $E$ of finite positive measure and every $\varepsilon \gt 0$ there exist a compact set $F \subset E$ and an open set $U \supset E$ such that $m(U \smallsetminus F) \lt \varepsilon$.

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:

For every integrable function $f$ and every $\varepsilon \gt 0$ there is a continuous function $\varphi$ such that $\int |f - \varphi| \lt \varepsilon$.


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.

Exercise.

  1. Give the details of the last sentence.
  2. Prove that for a Riemann integrable function the Lebesgue integral is defined and the Riemann and the Lebesgue integral coincide.
share|improve this answer
1  
Again, my profound gratitude. –  The Chaz 2.0 Nov 7 '11 at 1:46

I don't see any direct way to prove it from the theory provided, but it is quite easy to prove directly. (Since it is homework, I'm not going to spell it out completely)

Since f is integrable, then you have such a simple function - from the definition of the Lebesgue integral.

Now, from this simple function we can construct a step function, with an arbitrarily small integral of the difference. It is easy to do on a compact interval - you can split it enough so that you can't pay "too much" for the best fit (you use the fact that the function has a finite number of values). Now, you can combine a countable number of intervals by paying a diminishing amount for each (such as with $\frac{\epsilon}{2^n}$)

From the step function you can construct a continuous function - by connecting the various steps. By limiting the "width" of the connections, you can again make the difference integral as small as you wish (there is a countable number of jumps, so again $\frac{\epsilon}{2^n}$ is your friend here)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.