Suppose $f: [0,1] \to [0,1]$ is a continuous function. I want to show $$\mu(A)=\int_0^1 f(x)dx$$ where $A=\{(x,y)\in \mathbb{R}^2: y \le f(x)\}$ and $\mu(A)$ denotes the measure of $A$. In my proof I use $\mu^*(A)$ to denote the outer-measure of $A$.
My proof follows very closely to that of the general definition of the Riemann integral using upper and lower sums. I partition the horizontal axis into $n$ equal subintervals, and then create a family of elementary rectangles of width $1/n$ and height $\sup\{f(x):x \in [\frac{i-1}{n},\ \frac{i}{n}]\}$.
This set of elementary sets contains $A$, and thus: $\mu^*(A) \le \sum_{i=1}^n \frac{1}{n}\left(\sup\{f(x):x\in[(i-1)/n,\ i/n] \right)$
As $n \to \infty$ we have that $\mu^*(A) \le \overline{\int_0^1} f(x)dx$ which I use to denote the Riemann upper-sum.
Using similar notation and a similar method, I show $\underline{\int_0^1}f(x)dx \le \mu^*(A)$
Since $f$ is continuous and bounded, we know that the upper and lower sums are equal: $$\underline{\int_0^1}f(x)dx \le \mu^*(A) = \overline{\int_0^1} f(x)dx = \int_0^1 f(x)dx$$
And thus we have that $\mu^*(A)=\int_0^1 f(x)dx$.
Which seems nice and all $\textbf{except}$ that it assumes that $A$ is measurable at all.
If $A$ is measurable, then I can say that $\mu^*(A)=\mu(A)$, except that I don't know that $A$ is measurable at all. So, how can I go about proving that $A$ is measurable? I know that $A$ is measurable if $\mu^*(A)+\mu^*(A^c)=1$, but given an abstract set $A$ I don't really have a good way of going about identifying what $A^c$ is, much less proving the necessary condition to claim that $A$ is measurable.
So, my question is, can I simply claim that $\mu^*(A)=\mu(A)$? If so, why? Because that claim doesn't strike me as obvious in any way.
$A$ is a closed set by continuity of $f$ and every closed set is $\mu-$ measurable.
In fact, $A$ is a Borel set whenever $f$ is a Borel measurable function:
$A^{c}=\bigcup_{r \in \mathbb Q} [f^{-1}(-\infty,r) \times (r,\infty)]$.