The original question:
For any $f \in L_1[a,b]$, there exists a sequence of step function $h_n$ such that $\lim \int^a_b |h_n-f| = 0$
My approach is using littlewood principle, for any $\epsilon >0$, there exists a step function $g$ such that $ |f-g| < \epsilon$ except on a set A with measure lees than $\epsilon$.
I split the $\int^a_b |h_n-f|$ into $\int_{[a,b]\A} |h_n-f|$ and $\int_A |h_n-f|$, while the former one can be chosen to be less than any $\epsilon >0$ by suitable chosen $n$ , but I have no idea how to deal with $\int_A |h_n-f|$.
In addition, can I always choose a increasing sequence of step function which converges to any Lebesgue integrable function Any help would be appreciated.
First I will prove this for nonnegative integrable functions. The result follows directly from noting that $f=f^+ - f^-$, which are both nonnegative functions.
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.
Approximating simple functions by step functions is quite simple.