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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!

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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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  • $\begingroup$ Thanks very much for your answer. I will reread the proof again. However in Royden the definition of the lebesgue integral is the sup over the integrals of positive, bounded, finite support functions, and he makes a point of saying this is the definition for the integral whether or not the integral of f is finite. Therefore isn’t the fact that we can find $g$ bounded and finitely supported independent of the value of $\int f$? $\endgroup$ – P. Gillich Nov 1 '18 at 10:50
  • $\begingroup$ @P.Gillich: You're welcome. The definition of the integral allows $\int_E f = \infty$, however the problem states that $f$ is Lebesgue "integrable" which means the integral is finite. See the definition. If $\int_E f = +\infty$ then you can't write $\int_E g > \int_E f - \epsilon/2 = +\infty$. Note that $g$ is a bounded measurable function with finite support and so is Lebesgue integrable (i.e, $\int_E g < +\infty$). $\endgroup$ – RRL Nov 1 '18 at 16:44
  • $\begingroup$ That’s huge. Thanks! $\endgroup$ – P. Gillich Nov 2 '18 at 0:31

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