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!
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.