integrationmeasure-theoryreal-analysis
It can be proven that if a function, $f : [a,b] \rightarrow \mathbb{R}$ is Riemann-integrable then its graph is measurable and has Jordan measure $0$. Is the converse, for a function true?
That is;
If for a function $f : [a,b] \rightarrow \mathbb{R}$, the set $\{(x,f(x)) | x \in [a,b]\}$ is Jordan measurable with measure $0$, does it follow that $f$ is Riemann-integrable.
Let ${\mathcal D}(f)$ be the set of points at which $f$ is not continuous. The correct result is:
Theorem: Let $f:[a,b] \rightarrow {\mathbb R}$ be bounded. Then the following are equivalent:
(a) $\;f$ is Riemann integrable on $[a,b].$
(b) $\;{\mathcal D}(f)$ is a countable union of closed Jordan measure zero sets.
(c) $\;{\mathcal D}(f)$ is an ${\mathcal F}_{\sigma}$ Lebesgue measure zero set.
Nothing special is needed to prove this other than an examination of the usual proof characterizing the Riemann integrability of a function in terms of the zero measure of the set of points at which the oscillation of the function is greater than or equal to any specified positive real number. (Since these sets are closed, it doesn't matter whether we say Jordan measure or Lebesgue measure.) The countable union aspect shows up when we take the union of all such sets corresponding to the positive real numbers $\frac{1}{n}$ (as $n$ varies over the positive integers) in order to form the set of all points at which the function is not continuous.
A surprisingly little known consequence of this theorem is that the discontinuity set for any Reimann integrable function has both Lebesgue measure zero and is of the first (Baire) category (i.e. is meager). This is because any closed measure zero set (again, since "closed" is used, both Jordan and Lebesgue measures agree) is nowhere dense (a nearly trivial observation), and hence a countable union of closed measure zero sets is a first category set.
Incidentally, there exist sets that are both Lebesgue measure zero and first category (i.e. the sets are small in both the Lebesgue measure sense and the Baire category sense) which are too big to be covered by any discontinuity set of a Riemann integrable function, or even by the union of the discontinuity sets of countably many Riemann integrable functions. (I say "covered" because the issue is size, not Borel type. I'm disallowing someone from simply picking a non-${\mathcal F}_{\sigma}$ set, and letting this be the example by saying that the discontinuity set always has to be an ${\mathcal F}_{\sigma}$ set.) For more about this issue, specifically the fact that the $\sigma$-ideal of sets generated by the discontinuity sets of Riemann integrable functions is properly contained in the $\sigma$-ideal of sets that are simultaneously Lebesgue measure zero and first category, see my 30 April 2000 sci.math essay HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Bd}{\partial}\newcommand{\eps}{\varepsilon}$Hint: If your definition of Jordan measurable coincides with Wikipedia's (i.e., "inner" and "outer" approximations have arbitrarily close area), then each lower sum for the Riemann integral is an "inner" approximation and every upper sum is an "outer" approximation.
Edit: The strategy is to cover the graph of $f$ by finitely many rectangles $R_{i}$ of small area, then to subdivide $Q$ into rectangles $Q_{j}$ so that each $R_{i}$ is partitioned (modulo overlapping boundaries) by its intersections with the "columns" $Q_{j} \times [0, M]$. Finally, over each $Q_{j}$, choose an interval $[m_{j}, M_{j}]$ so that
$m_{j} \leq f(x) \leq M_{j}$ for all $x$ in $Q_{j}$, and
$Q_{j} \times [m_{j}, M_{j}]$ is contained in the union of the $R_{i}$.
The corresponding approximations to the Riemann integral are arbitrarily close.
Since the details are potentially delicate, here's a sketch of the one-dimensional case, phrased in a way to facilitate generalization to arbitrarily many variables.
Let $Q$ be a closed interval, $f:Q \to \Reals$ a bounded function satisfying $0 \leq f(x) \leq M$ for all $x$, and $$ \Gamma = \{(x, y) : 0 \leq f(x) \leq y,\ x \in Q\} $$ the Jordan-measurable region under the graph of $f$.
Fix $\eps > 0$, and pick finitely many rectangles $R_{i} = [a_{i}, b_{i}] \times [c_{i}, d_{i}]$ (with $1 \leq i \leq I$, say) of total area at most $\eps/2$ whose union contains the graph $G$ of $f$. (Use the remaining $\eps/2$ of the "total error budget" to cover $(Q \times \{0\}) \cup (\Bd Q \times [0, M])$.) There exists a partition $P = (x_{j})_{j=0}^{J}$ of $Q$ such that for every $j = 1, \dots, J$ and for every $i = 1, \dots, I$, either $Q_{j} = [x_{j-1}, x_{j}]$ is contained in $[a_{i}, b_{i}]$, or else $(x_{j-1}, x_{j}) \cap (a_{i}, b_{i}) = \varnothing$. [Note 1]
For each $j$, let $m_{j}$ be the largest of the $c_{i}$ such that
$Q_{j} \subset [a_{i}, b_{i}]$, and
$c_{i} \leq f(x)$ for all $x$ in $Q_{j}$.
In words, $m_{j}$ is the height of the highest lower edge among rectangles lying over $Q_{j}$ that also lies on or below the graph of $f$ over $Q_{j}$.
Similarly, let $M_{j}$ be the smallest of the $d_{i}$ such that $Q_{j} \subset [a_{i}, b_{i}]$ and $f(x) \leq d_{i}$ for all $x$ in $Q_{j}$. [Note 2]
Consequently, [Note 3] $$ U(f, P) - L(f, P) = \sum_{j=1}^{N} (M_{j} - m_{j})(x_{j} - x_{j-1}) \leq \sum_{i=1}^{I} (d_{i} - c_{i})(b_{i} - a_{i}) < \eps/2. \tag{*} $$ In words, the Darboux sums can be made closer than $\eps/2$ for arbitrary $\eps > 0$, so $f$ is Riemann integrable over $Q$.
Notes:
In the one-dimensional case, take $P = (x_{j})$ to be the sorted union of the division points $a_{i}$ and $b_{i}$. Generally, subdivide this way "in each variable". Geometrically, cover the graph of $f$ with finitely many boxes $R_{i}$ of total volume at most $\eps/2$. Project away the $(n+1)$th coordinate, and take the smallest grid subdivision of $Q$ that contains the shadows of the $R_{i}$.
Each $Q_{j}$ is contained in at least one $[a_{i}, b_{i}]$, and by construction is completely contained in every $[a_{i}, b_{i}]$ whose interior intersects $Q_{j}$. Because the $R_{i} = [a_{i}, b_{i}] \times [c_{i}, d_{i}]$ cover the graph of $f$, for each $j$, there exists an $i$ such that $c_{i} \leq f(x)$ for all $x$ in $Q_{j}$, and there exists an $i'$ such that $f(x) \leq d_{i'}$ for all $x$ in $Q_{j}$.
By construction, for each $j = 1, \dots, J$ we have $m_{j} \leq f(x) \leq M_{j}$ for all $x$ in $Q_{j}$. The first equality in (*) follows since the $Q_{j}$ have disjoint interiors.
By construction, $Q_{j} \times [m_{j}, M_{j}] \subset \bigcup_{i=1}^{I} [a_{i}, b_{i}] \times [c_{i}, d_{i}]$, so $$ \bigcup_{j=1}^{J} Q_{j} \times [m_{j}, M_{j}] \subset \bigcup_{i=1}^{I} [a_{i}, b_{i}] \times [c_{i}, d_{i}], $$ which gives the first inequality in (*).
Best Answer
Take the function $f :[0,1] \to \mathbb{R}$ where
$f(x)=\begin{cases}\ 1 & x \in \mathbb{Q}\cap [0,1]\\ 0 & x \in \mathbb{Q}^c \cap [0,1] \end{cases}$
The graph of this function has Jordan content zero which implies that it has Jordan measure zero.
But $f$ is not Riemman integrable.