Area function proof from Apostol Vol 1 Calculus

Question

I was reading Apostol's Calculus Volume 1, on Area Under a Curve. Apostol has claimed (intuitively true of course!) that the GRAPH of the function has no area. More rigorously, define the ordinate set of an integrable function $f$ as $Q=\{(x,y)|0\leq y\leq f(x),a\leq x\leq b\}$ where $f$ is defined on the closed, bounded interval $[a,b]$.

Then, $Q$ is measurable in the sense that a positive number called $area$,noted as $a$, can be assigned to $Q$ and $a(Q)=\int_a^bf(x)dx$.

If we define $Q'=\{(x,y)|0\leq y<f(x),a\leq x\leq b\}$ (note the difference between $Q$ and $Q'$) then $Q'$ is measurable and $a(Q)=a(Q')$. This is Apostol's claim and intuitively this is true. But it is not immediately clear to me.

Apostol has left without proving it by simply saying that the proof is similar to the way in which we prove $a(Q)=\int_a^bf(x)dx$. But I fail to see any rigorous interconnection between those two.

Answer 1

Set $\forall n: Q_n:=\{(x,y)|x\in[a,b],y\in[0,f(x)-1/n]\}$. Then $\lim Q_n=Q'$. We are given the definition of the area including the graph of the function, namely the integral of the function. As $f$ is integrable, so is $\forall n: g_n(x):=f(x)-1/n$. Then we have $a(Q_n)=a(Q)-(b-a)/n$. Taking the limit of both sides we have the desired equality.

Note: For simplicity we may assume $f>1$.

Answer 2

1. $S \subseteq Q' \subseteq T $.
2. Because $f$ is integrable, there is exactly one number, called $I$, such that $a(S) \leq I \leq a(T)$.
3. Because of $1$ and $2$ and because of the exhaustion property, it follows that $Q'$ is measurable and $a(Q') = I$.
4. Because $a(Q) = I \land a(Q') = I \implies a(Q) = a(Q')$