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

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!

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.

• 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 $\int f$? – P. Gillich Nov 1 '18 at 10:50
• @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, $\int_E g < +\infty$). – RRL Nov 1 '18 at 16:44
• That’s huge. Thanks! – P. Gillich Nov 2 '18 at 0:31