[Math] Show that for a positive integrable function $f$ there is a simple function $\eta$ such that $0 \leq \eta \leq f$ and $\int |f-\eta|<\epsilon$

lebesgue-integrallebesgue-measure

This is a problem from Royden & Fitzpatrick that has been giving me trouble for a couple of days. I feel like I'm close to a solution but there are a few points that make me unsure. For those who might have the book, the problem is # 21 in chapter 4. It goes like this:

  1. Let $f \geq 0$ be a function integrable over the measurable set $E$. Let $\epsilon>0$. Show that there exists a simple function $\eta$ on $E$ that has finite support, $0 \leq \eta \leq f$ on $E$, and $\int_E |f – \eta| <\epsilon$.

  2. If $E$ is closed and bounded, show that there is a step function $h$ on $E$ with finite support and $\int_E |f – h| <\epsilon$.

Attempt at a solution:

$\int_E f = \sup \{\int_E g : \text{$g$ is bounded, with finite support, $g \geq 0 $ on E}\}$. This implies that

$\exists g$ for which $$\int_E g > \int_Ef – \frac{\epsilon}{2}.$$
$$\Rightarrow \int_E (f-g) = \int_E |(f-g)| <\frac{\epsilon}{2} $$

Since $g \leq f$ on $E$. By the simple approximation theorem, we can approximate $g$ by an increasing sequence of simple functions $\{\phi_n\}$ such that $\phi_n = 0$ whenever $g =0.$ In particular this means that $\phi$ has finite support as well.

Now, here is where I'm having trouble. What is the guarantee that $\phi_n$ is eventually $\geq 0$ on $E$? I understand that this sequence of simple functions has to converge pointwise to $g\geq 0 $ on $E$, but this doesn't guarantee that for some single $n$ the function $\phi_n$ is uniformly positive over the whole domain $E$.

However, let's assume that it's true that I can find such a sequence of simple functions: then since $g$ is bounded, by the bounded convergence theorem

$$\lim_{n\rightarrow \infty}\int_E \phi_n = \int_E g.$$

Now this means that for $N$ large enough, $\int_E \phi_N + \frac{\epsilon}{2}> \int_E g$. Let $\eta = \phi_N$. In particular, as before $\int_E |(g-\eta)| <\frac{\epsilon}{2}$

Finally, we get the estimate

$\int_E |f – \eta| \leq \int_E (|f-g|+|g-\eta|) = \int_E |f-g| + \int_E |g-\eta|< 2\frac{\epsilon}{2} = \epsilon$.

My concerns are:

  1. What I already mentioned about the sequence of simple functions being positive

  2. I don't see where the integrability of the function $f$ comes in here. According to the definition, it means that $f$ just has finite Lebesgue integral. I know all of what I've just written must be garbage if I don't use all the hypotheses of the question.

  3. I don't have the faintest clue how to proceed with the second part of the question.

Any help you all would be able to give would be greatly appreciated.

Thanks!

Best Answer

For Question 2

The Lebesgue integrability of nonnegative, measurable $f$ means $\int_Ef < +\infty$.

It is then true by the definition of the integral that there exists a bounded, measurable function $g$ of finite support such that $0 \leqslant g \leqslant f$ and

$$\tag{1}\int_E |f-g| < \frac{\epsilon}{2}$$

For Question 1

Let $E_0 = \text{supp }(g)$. It only remains to produce a simple function $\eta$ with finite support in $E_0$ such that $0 \leqslant \eta \leqslant g$ and

$$\tag{2}\int_{E} |g-\eta| = \int_{E_0} |g-\eta| < \frac{\epsilon}{2}$$

The Simple Approximation Theorem (along with the Dominated Convergence Theorem) gives you everything you need to prove (2). Since $g$ is nonnegative with finite support, there exists an increasing sequence of simple functions $\{\phi_n\}$ with finite support such that $0 \leqslant \phi_n \leqslant g$ and $\phi_n \to g$ pointwise. Reread the statement including special cases and proof of this theorem in Royden.

Since $g$ is integrable, by the Dominated Convergence Theorem we have

$$\lim_{n \to \infty} \int_{E_0} \phi_n = \int_{E_0} g$$

Given $\epsilon > 0$ there exists $N $ such that

$$\int_{E_0} |g - \phi_N| = \int_{E_0} (g - \phi_N)= \int_{E_0}g - \int_{E_0}\phi_N < \frac{\epsilon}{2}$$

Taking $\eta = \phi_N$ proves (2).

For Question 3

It would be better to post this as another question, but here is a sketch.

Since this problem arises in Ch.4 we can assume $E \subset \mathbb{R}$.

It is enough to prove that for a characteristic function $\chi_A$ on a measurable set $A \subset E$, there is a step function $\phi$ such that $\int_E| \chi_A - \phi| < \epsilon$. For any $\delta > 0$, there is an open set $O$ containing $A$ with $m(O \setminus A) < \delta$. As an open set $O$ is a countable union of disjoint open intervals

$$O = \bigcup_{j=1}^\infty(a_j,b_j),$$

we have

$$m(O) = \sum_{j=1}^\infty(b_j-a_j) < m(A) + \delta$$

Construct $\phi$ as the characteristic function of a finite union of a sufficiently large number of the intervals $(a_1,b_1), \ldots ,(a_m,b_m)$. This is a step function with the desired properties.

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