Area under zero almost everywhere step function

Question

I feel like this is a basic question but I am having trouble getting rigorous. Given $f$ a nonnegative step function, hence we can write $f=\sum^{n}_{i=1}a_i \chi_{A_i}$ where $A_i's$ are disjoint measurable subsets and $f$ takes $a_i$ on $A_i$. Define $A(f)=\sum^{n}_{i=1}a_i m(A_i)$. If $f=0$ almost everywhere, then $A(f)=0$.

My attempt:

Let $N$ be the null set such that $f(x)=0$ for each $x\in N^c$. How do I continue? I am thinking of setting $A_i=(A_i \cap N)\cup (A_i\cap N^c)$, but I don't know how to manipulate $\chi_{A_i}$.

Any suggestion is appreciated. Thanks a lot.

Answer 1

If exists $m \in \{1,2,...,n\}$ and $a_m>0$ then we must have $m(A_m)=0$ otherwise we would have that $f(x)>0\forall, x \in A_m$ which contradicts hypothesis.

So in every case we have that every term of the linear combination has the form $0 \cdot a_i$ or $0 \cdot m(A_i)$ which are both zero.(Of course we have the agreement that $0 \cdot \infty=0$)

So $A(f)=0$