# Simple function with same support as a function with finite support

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.