# The Unapologetic Mathematician

## The Measures of Ordinate Sets

If $(X,\mathcal{S})$ is a $\sigma$-algebra and $f:X\to\mathbb{R}$ is a Borel-measurable function we defined the upper and lower ordinate sets $V^*(f)$ and $V_*(f)$ to be measurable subsets of $X\times\mathbb{R}$. Now if we have a measure $\mu$ on $X$ and Lebesgue measure on the Borel sets, we can define the product measure $\lambda$ on $X\times\mathbb{R}$. Since we know $V^*(f)$ and $V_*(f)$ are both measurable, we can investigate their measures. I assert that

$\displaystyle\lambda(V^*(f))=\int f\,d\mu=\lambda(V_*(f))$

It will be sufficient to establish this for simple functions, since for either the upper or the lower ordinate set we can approximate any measurable $f$ by a monotone sequence of simple $\{f_n\}$ so that $\lim\limits_{n\to\infty}V^*(f_n)=V^*(f)$ or $\lim\limits_{n\to\infty}V_*(f_n)=V_*(f)$. Then the limit will commute with $\lambda$ (since measures are continuous), and it will commute with the integral as well.

So, we can assume that $f$ is simple, and write

$\displaystyle f=\sum\limits_{i=1}^n\alpha_i\chi_{E_i}$

with the $\{E_i\}$ a pairwise-disjoint collection of measurable sets. But now if the equality holds for each of the summands then it holds for the whole function. That is, we can assume — without loss of generality — that $f=\alpha\chi_E$ for some real number $\alpha$ and some measurable subset $E\subseteq X$.

And now the result should be obvious! Indeed, $V^*(f)$ is the measurable rectangle $E\times[0,\alpha]$, while $V_*(f)$ is the measurable rectangle $E\times[0,\alpha)$. Since the product measure on a measurable rectangle is the product of the measures of the two sides, these both have measure $\alpha\mu(E)$. On the other hand, we calculate the integral as

$\displaystyle\int f\,d\mu=\int\alpha\chi_E\,d\mu=\alpha\mu(E)$

and so the equality holds for such rectangles, and thus for simple functions, and thus for all measurable functions.

Of course, it should now be clear that the graph of $f$ has measure zero. Indeed, we find that

$\displaystyle\lambda(V^*(f)\setminus V_*(f))=\lambda(V^*(f))-\lambda(V_*(f))=\int f\,d\mu-\int f\,d\mu=0$

These results put a precise definition to the concept of the integral as the “area under the graph”, which was the motivation behind our definition of the Riemann integral, way back when we introduced it.