# Does the Lebesgue's criterion for Riemann-integrability also hold for Riemann-Stieltjes integral?

Lebesgue's criterion for Riemann-integrability:

Let $f$ be defined and bounded on $[a,b]$ and let $D$ denote the set of discontinuities of $f$ in $[a,b]$.Then $f\in R$ on $[a,b]$ if,and only if,$D$ has measure zero.

My question is,does the Lebesgue's criterion for Riemann-integrability also hold for Riemann-Stieltjes integral?That is,does the following hold?

Let $f$ be defined and bounded on $[a,b]$ and let $D$ denote the set of discontinuities of $f$ in $[a,b]$.Then $f\in \mathcal{R}(\alpha)$ on $[a,b]$ if,and only if,$D$ has measure zero.($\alpha$ is a monotonically increasing function on $[a,b]$).

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In the case of R.-S., you want to replace measure zero with countable. For details, see here: en.wikipedia.org/wiki/Riemann_integral. – William Jun 22 '12 at 4:01
@William Are you sure that countable gives "if and only if"? Certainly not "only if", because $\alpha(x)$ could be $x$, for example. And not "if" either, because if $f$ is discontinuous just at one point which happens to have positive mass, the integral does not exist. – user31373 Jun 22 '12 at 16:03
@LeonidKovalev: Perhaps a better conjecture would be that $f \in \mathcal{R}(\alpha)$ iff $D$ has $\alpha$-measure zero. Of course, one must correct this based on your discussion below about one-sided continuity. – Nate Eldredge Jun 22 '12 at 16:40

Chances are that we are using the definition of R-S integral as found in Rudin's Principle of Mathematical Analysis. That is, the upper R-S sum is $U(P,f,\alpha)=\sum_{i=1}^nM_i (\alpha(x_i)-\alpha(x_{i-1}))$ where $M_i=\sup\{f(x): x\in [x_{i-1},x_i]\}$. The upper integral is the infimum of all upper sums. The lower integral is defined similarly, and when both are equal, we declare the function integrable wrt $\alpha$.
Confession: I hate the above definition. It leads to ridiculous hair-splitting: the functions $$\alpha_1(t)=\begin{cases} 1,\qquad t>0 \\ 0,\qquad t\le 0\end{cases}$$ $$\alpha_2(t)=\begin{cases} 1,\qquad t\ge 0 \\ 0,\qquad t< 0\end{cases}$$ $$\alpha_3(t)=\frac{1}{2}(\alpha_1+\alpha_2)$$ are all different ways to assign unit mass to the point at $0$. The first puts all mass on the "right half of the point", the second puts it all on the "left half", and the third gives $1/2$ to each half of the point.
So, a function $g$ is integrable with respect to $\alpha_1$ iff $g(0)=g(0+)$ - it does not have to be continuous from the left at $0$. Similarly, it is integrable wrt $\alpha_2$ iff $g(0)=g(0-)$. In both of these examples the set of discontinuities may well be all of $\mathbb R$, yet we have integrability as long as one-sided continuity at $0$ holds. Thus, integrability cannot be expressed in terms of the set of discontinuity $D$: you have to look at the set of right discontinuities $D_+$ and left discontinuities $D_-$. And it's not about measure in the usual sense, because we are "measuring" half-points.
Honestly, I don't know why people do this to themselves instead of assuming one-sided continuity of $\alpha$ and defining $M_i$ in terms of half-open intervals.