1
$\begingroup$

I am looking at a solution to the first part of problem 21 in Section 4.3 of Royden's Real Analysis.

The part of the problem I am interested in states as follows:

Let the function $f$be nonnegative and integrable over $E$ and $\epsilon > 0$. Show tat there is a simple function $\eta$ on $E$ that has finite support, $0 \leq \eta \leq f$ on $E$ and $\int_{E}|f-\eta|< \epsilon$.

I am trying to justify the steps of the solution to myself.

The solution starts off as follows:

Let $\epsilon > 0$. We know by the definition of the integral for non-negative measurable functions that there exists a bounded, measurable function of finite support, $h \leq f$, such that $\int_{E} h > \int_{E}f-\epsilon/2$.

Up to this point, I am fine. It then goes on to state:

We also know that for any given bounded, measurable function $h$ with finite support, there exists a simple function $\eta \leq h$ with the same support as $h$ such that $\int_{E}\eta > \int_{E}h - \epsilon/2$.

At the beginning of this section in Royden, he defines the integral of a nonnegative measurable function $f$ over $E$ by $\int_{E}f = \sup \left\{\int_{E}h\vert h\,\text{bounded, measurable, of finite support, and}\, 0 \leq h \leq f\,\text{on}\,E\right\}$. But, this is the only thing even remotely close to saying you can find a function g with finite support such that $\int_{E} g > \int_{E}f - \epsilon/2$. How, if $f$ already has finite support, then, do we know that we can find a function $h$ that not only has finite support, it has the same support as $f$ and is simple?!

I would like a full proof in your answer - nothing too complicated; just using what is available to us up to Section 4.2 of Royden. Thank you.

$\endgroup$
1
$\begingroup$

So you are asking if this definition for integral of non-negative function is well-defined? If so, by The Simple Approximation Theorem,page 62, measurable function can be approximated by simple function, and since the definition take the supermum, and combine with comment on his book : enter image description here

So it is well-defined.

By the way, in Royden's book, page 85, problem 24 gives another definition of integral of non-negative measurable functions.

Here is a solution for general case PDF.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy