Measure Theory/Integrable Almost Continuous

Integrable Is Almost Continuous

In this lesson we will prove that if $f:ℝ\to ℝ$ is integrable then for every $\epsilon \in {ℝ}^{+}$ there is a continuous function $g:ℝ\to ℝ$ such that

$\int |f-g|<\epsilon$

Integrable Is Approximately Simple

Going directly toward that goal doesn't have an immediately clear path. Instead, we set a simpler goal, literally. We start by showing the corresponding result for simple functions, rather than continuous functions.

That is to say, here we will show that under the same conditions, there is a simple $\psi :ℝ\to ℝ$ such that

$\int |f-\psi |<\epsilon$

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Because we assume f is integrable, as a general measurable function, we need to consider two integrals, $\int f^{+}$ and $\int f^{-}$. We know that each of these is finite and nonnegative.

By their definitions as nonnegative integrals, there exist simple functions such that $0\le \phi \le f^{+}$ and $0\le \psi \le f^{-}$ and

$\int (f^{+}-\phi ),\int (f^{-}-\psi )<\epsilon$

Finish the proof started above. In particular, show that there is a simple function $\chi$, such that $\int |f-\chi |$ is arbitrarily small.

(I am being deliberately coy about what the function $\chi$ is. It must be somehow related to $\phi ,\psi$ but I want you to think about exactly how.)

Integrable is Approximately Step

Exercise 3. Approximate Step Integration is the focus of this subsection. However, we will need the result of Exercise 2. Restricting the Domain Makes Integrals Small will be helpful for us to complete Exercise 3.

Let $f:ℝ\to ℝ$ be an integrable function and $\epsilon \in {ℝ}^{+}$ arbitrary.

Show that there exists a $\delta \in {ℝ}^{+}$ such that for every measurable subset $E\subseteq ℝ$ with $\lambda (E)<\delta$ we have

$\underset{E}{\int }f<\epsilon$

Hint: Use $\epsilon$ and the monotone convergence theorem to find a simple function $0\le \psi \le f$ such that $\int f<\epsilon +\int \psi$.

Now select $\delta$ small enough to make $\psi \delta <\epsilon$.

Use these ingredients to split $\int f$ into $\int (g-\psi )+\int \psi$ and proceed with the properties of integrals.

Use the earlier result relating simple to step functions, to show that there is a step function s such that

$\int |f-s|<\epsilon$

with f and $\epsilon$ as before.

Hint: Split the integral into two domains, one of which has a small domain and the other has a small integrand. Show that both terms are small.

Integrable Is Approximately Continuous

With the assumptions as before, now show that there is a continuous function g such that $\int |f-g|$ is small. Given what we've shown above, at some point in your solution you should observe that for any simple function $\psi$,

$\int |f-g|\le \int |f-\psi |+\int |\psi -g|$

and the left-hand term can be chosen to be small. After making this choice, you should be able to then make a choice of g which makes the second term small.