I have the following theorem from my class notes for Stochastic Analysis.
If $f\in L^1$ there exists a sequence of simple functions $f_n$ such that $f_n$ converges to $f$ in $L^1$.
For the proof, there's only the description of the usual way to construct simple functions $f_n$ approximating $f\ge 0$ by setting $f_n(w)=2^n \cdot 1_{f^{-1}[2^n, \infty]}(w)+\sum_0^2{^{2^n-1}} \frac{k}{2^n} \cdot 1_{f^{-1}([\frac{k}{2^n}, \frac{k+1}{2^n})}$. I can see that for integration with respect probability measure these $f_n$ do the job, but for arbitrary measures does this result still hold? There's not a reference to the measure in the theorem so I'm confused if this result is general or not. I would greatly appreciate any help.
I think this should work, I am a little out of practice with measure theory. \
By definition of $f \in L^1$ and nonnegative $\exists h \leq f$ bounded, measurable, and finite support such that $\int{|f-h|} < \epsilon / 2$. Now, as h is a measurable function $\exists \{\phi_n\} $ simple such that $|\phi_n | \leq |h|$ for all n and since h is non-negative we can choose the sequence to be increasing and $\phi_n \to h$. Then we have that $\int{|f - \phi_n |} \leq \int{|f-h|} + \int{|h- \phi_n|} < \epsilon / 2 +\int{|h- \phi_n |}$. Now, we have that $|h - \phi_n| \leq 2 |h|$ and $2 |h| \in L^1$ so applying the Lebesgue Dominated Convergence we can justify integral limit swap to conclude that $\int{|h- \phi_n |} < \epsilon /2$. Giving us a general result.