Approximate an integral with a step function 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!
 A: 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:

Lemma (Exercise 3.2.16): Let $I\subseteq\mathbb{R}$ be a compact interval, $s:I\to\mathbb{R}$ be simple. Then $\forall\varepsilon>0,\exists$ step $\varphi:I\to\mathbb{R},\exists F\in\mathcal{M}:$
\begin{align}
&F\subseteq I, \forall x\in F: \varphi(x)=s(x),\\
&m(I-F)<\varepsilon.
\end{align}
Moreover we can bound $\varphi$ on $I$ by $M:=\Sigma_{i=1}^n|\alpha_i|$ provided that $s=\Sigma_{i=1}^n \alpha_i \chi_{E_i}$ is the canonical representation of $s$.

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:

Lemma (Exercise 4.3.21): Let $I\subseteq\mathbb{R}$ be a compact interval, $f:I\to[0,\infty]$ be integrable. Then $\forall\varepsilon>0,\exists$ step $\varphi:I\to\mathbb{R}:m(\{\varphi\neq0\})<\infty,\int_E|f-\varphi|<\varepsilon$.


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.
A: First,
don't worry about
the coefficient of
$\epsilon$.
Just make it $1$
everywhere you can
and add up all the coefficients.
As long as you come up with
$\epsilon$
multiplied by some constant,
you are fine.
In your case,
start with
$\int_\mathbb{R} |f-\eta|<\epsilon
$.
Next,
suppose that
$|\eta - s|<\epsilon
$.
Then
${\int_{[-N,N]} |\eta-s|}
<2N\epsilon
$.
Adding up all these,
you get an error
of less than
$(2N+2)\epsilon
$.
If you don't like this,
in the
$|\eta - s|<\epsilon
$
above,
replace it by
$|\eta - s|<\frac{\epsilon}{2N}
$.
This will make your total error
$3\epsilon
$.
I always make my initial bound
$\epsilon$.
Then,
I see what the total
error is.
