# Riemann Stieltjes Integral of discontinuous function

A function $f(x)$ on $[a,b]$ with finite number of discontinuities is Riemann-Stieltjes integrable if $\alpha$ is continuous where $f$ is discontinuous.In proof we use continuity of $\alpha$ .Can we find an example for a function $f$ and a monotonically increasing function $\alpha$ such that $\alpha$ is discontinuous at some points of discontinuity of $f$ to show that integrability of $f$ fails ?

## 1 Answer

Choose $f=\alpha = 1_{[{1 \over 2},1]}$ on $[0,1]$.

Let $P_n = (0, {1 \over 2n}, {2 \over 2n},..., 1)$ (so ${1 \over 2} \in P_n$ for all $n$). Let $Q_n$ be $P_n$ with the point ${1 \over 2}$ removed). Note that the mesh size of the partition $Q_n$ is ${1 \over n}$, in particular, it is arbitrarily small.

If we choose $c_k$ such that $c_k \in [a,b]$ where $a,b$ are consecutive points in $Q_n$, the we have the sum $S(Q_n, \{c_k\}, f, \alpha) = f(c_n) (\alpha({n+1 \over 2n})-\alpha({n-1 \over 2n})) = f(c_n)$, where $c_n \in [{n-1 \over 2n}, {n+1 \over 2n}]$. Hence by choosing $c_n$ appropriately (or inappropriately as is the case here) the sum can be either zero or one regardless of mesh size. Hence no limit can exist and $f$ is not integrable with respect to $\alpha$.

• Can we say that $f=1$ discontinuous in [0,1] ? – Madhu Oct 20 '14 at 2:29
• I don't understand your question. – copper.hat Oct 20 '14 at 2:33
• Theorem says that even though $f$ has finite points of discontinuities $f$ is Riemann -Stieltjes integrable if $\alpha$ is continuous at the points of discontinuities of $f$.Before moving to proof I want to verify that continuity of $\alpha$ is necessary using an example.I mean I want to verify with an example that a discontinuous function $f$ along with a discontinuous but Monotonically increasing function $\alpha$ is not Riemann Stieltjes integrable – Madhu Oct 26 '14 at 15:19
• I'm really not sure what you are looking for. The $f$ I gave above is discontinuous, as is $\alpha$. Both are non-decreasing. You can add something like $x \mapsto x$ to both to make them strictly increasing, but the net result will be the same. – copper.hat Oct 27 '14 at 2:31