First of all, I want to point out that I know certain things. These are tools that I may use:
(i). If $f$ is integrable over $\mathbb{R}$ then there is a simple function $\eta$ such that $\eta$ has finite support and $\int_\mathbb{R} |f-\eta|<\epsilon$.
(ii). Given a measurable function $\eta$ on a closed and bounded interval $I$, there exists a step function $s$ on $I$ and measurable set $F\subset I$ so that $|\eta -s|<\epsilon$ on $F$ and $m(I\setminus F)<\epsilon$.
Claim: If $f$ is integrable over $\mathbb{R}$ then There exists a step function $s$ which vanishes outside of a closed interval and $\int_\mathbb{R} |f-s|<\epsilon$.
Proof(attempt): First of all, since $f$ is in fact integrable on all of $\mathbb{R}$ then most of its mass is contained in some bounded interval, hence let $N\in\mathbb{N}$ so that $\int_{\mathbb{R}\setminus[-N,N]} |f| <\epsilon/3$. Now apply (i) to obtain the simple $\eta$ so that $\int_\mathbb{R} |f-\eta|<\epsilon/3$. Since $\eta$ is measurable and defined on $[-N,N]$, by (ii) let step function $s$ exist on $[-N,N]$, vanishing outside of it, so that $|\eta – s|<(?)$ except on a set of measure (?). Then we have:
\begin{align*}
\int_\mathbb{R}|f-s|&=\int_{\mathbb{R}\setminus[-N,N]}|f| + \int_{[-N,N]} |f-s|\\
&=\int_{\mathbb{R}\setminus[-N,N]}|f| + \int_{[-N,N]} |f-\eta+\eta-s|\\
&\leq\int_{\mathbb{R}\setminus[-N,N]}|f| + \int_{[-N,N]} |f-\eta|+ \int_{[-N,N]} |\eta-s|\\
&<\epsilon/3 + \epsilon/3 + \underbrace{\int_{[-N,N]} |\eta-s|}_{\text{I need help controlling this}}\\
\end{align*}
Any suggestions? I feel like I'm either very close, or the approach was wrong entirely. Thanks!
Best Answer
Let me first give accurate references to make citations less convoluted. The result you are citing is Exercise 4.6.44.(ii) from Royden & Fitzpatrick's Real Analysis, 4e (p. 96), while your lemmas (i), (ii) are Exercises 4.6.44.(i) (p. 95) and 3.2.18 (p. 63). Although to use these two results is suggested as a hint in the book, I believe there are less painful ways to do this approximation. I'll outline two of them.
1st way: First we prove this for $f:\mathbb{R}\to[0,\infty]$. Let $\varepsilon>0$. Like you started, first cut off the integral so that $$\int_{\mathbb{R}-[-N,N]}f<\dfrac{\varepsilon}{2}$$
and then do an approximation by a simple function $s:[-N,N]\to[0,\infty[$ using 4.6.44.(i):
$$\int_{[-N,N]}|f-s|<\dfrac{\varepsilon}{4}.$$
Next, instead of weakening $s$ to be a measurable function, we consider it as is, and use Exercise 3.2.16 (p. 63) instead of 3.2.17:
I added the last statement in the lemma, though it is immediate from the very construction of the required step function. Also observe that this answers your problem: we can have a bound on the integral over the compact interval that is dependent only on the values taken by the simple function $s$. (Consequently you were very close indeed.)
If we use the lemma to approximate $s$ by a step function with error $\dfrac{\varepsilon}{8M}$, $\int_{[-N,N]}|s-\varphi|<\dfrac{\varepsilon}{4}$. Noting that we can consider $\varphi$ to have the real line as its domain by $\varphi \chi_{[-N,N]}:\mathbb{R}\to\mathbb{R}$, we are done for nonnegative $f$.
For the general case it suffices to apply this argument to the positive and negative parts of $f$ (starting with half the error); everything adds up nicely.
2nd Way: Another way is to apply Exercise 4.3.21 (p. 85) directly to the positive and negative parts of $f$, after cutting off the integral outside a compact interval:
P.S. To be sure the first argument can be used to prove Exercise 4.3.21 so that I am really suggesting one solution only. That said I believe all these multilinks between exercises are mainly due to the step-by-step approach of the book.