I am currently reading through chapter 11 of Rudin's Principles of Mathematical Analysis, and I'm trying to solve problem 7:
Find a necessary and sufficient condition that $f \in \mathfrak R(\alpha)$ on $[a,b]$. [the class of Rienmann-Stieltjes integrable functions, with integrator $\alpha$]
Hint: Consider Example 11.6(b) and Theorem 11.33.
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In example 11.6(b) Rudin defines an additive set function on
intervals according to a monotonically increasing function $\alpha$:$$\mu((a,b))=\alpha(b^- )-\alpha(a^+)\\
\mu((a,b])=\alpha(b^+)-\alpha(a^+)\\
\mu([a,b))=\alpha(b^-)-\alpha(a^-) \\
\mu([a,b])=\alpha(b^+)-\alpha(a^-) $$and states that it can be extended to a countably-additive set
function over a richer sigma-algebra. - In Theorem 11.33 he proves that any Rieamann integrable function is Lebesgue integrable (dm) and that their values coincide. He also proves the criterion for Riemann integrability – continuity almost everywhere (dm). The method of the proof relies on Darboux sums.
I'd greatly appreciate any help.
EDIT: This is my attempt at the solution. Please tell me what do you think about it.
First of all we extend $\alpha:[a,b] \to \mathbb R$ to a monotonically increasing function over $\mathbb R$, as follows: for $x<a$, $\alpha(x):=\alpha(a)$ and for $x>b$, $\alpha(x):=\alpha(b)$. We now can use example 11.6(b) to get a measure $\mu_\alpha$ over a sigma-algebra of subsets of $\mathbb R$, and we will immediately restrict it back to the interval $[a,b]$.
I will try to prove that $f \in \mathfrak R(\alpha) \Leftrightarrow$ $f$ is continuous a.e. ($d \mu_\alpha$)
Suppose $f$ is bounded. By definition 6.1 and theorem 6.4 there is a sequence $\{P_k \}$, of partitions of $[a,b]$, such that $P_{k+1}$ is a refinement of $P_k$, such that the distance between adjacent points of $P_k$ is less than $\frac{1}{k}$, and such that: $$\lim_{k \to \infty} L(P_k,f,\alpha)=\underline{\int}_a^b f d\alpha, \, \lim_{k \to \infty} U(P_k,f,\alpha)= \overline{\int}_a^b f d\alpha $$ Furthermore, since $\alpha$ is monotonic, it has only countably many points of discontinuity, so we can take the points of the partitions to be points where $\alpha$ is continuous.
If $P_k=\{x_0,x_1,\ldots,x_n\}$ with $x_0=a,x_n=b$, define the functions $L_k,U_k:[a,b] \to \mathbb R$ according to: $$L_k(x)=m_1,U_k(x)=M_1$$ for $a \leq x \leq x_1$ and $$L_k(x)=m_i,U_k(x)=M_i $$ for $x_{i-1} < x \leq x_i,2 \leq i \leq n$.
Notice that: $$\int_{[a,b]} L_k d\mu_\alpha=L(P_k,f,\alpha), \, \int_{[a,b]} U_k d\mu_\alpha=U(P_k,f,\alpha)$$ due to the continuity of $\alpha$ at the points of the partitions $\{ P_k \}$.
We also have $$L_1(x) \leq L_2(x) \leq \ldots \leq f(x) \leq \ldots \leq U_2(x) \leq U_1(x) $$ for all $x \in [a,b]$, since $P_{k+1}$ refines $P_k$. We therefore have the existence of $$L(x):=\lim_{k \to \infty} L_k(x),U(x):= \lim_{k \to \infty} U_k(x) $$ and $L,U$ are bounded and measurable on $[a,b]$, $L(x) \leq f(x) \leq U(x)$ and $$\int_{[a,b]} L d\mu_\alpha=\underline{\int}_a^b f d\alpha, \, \int_{[a,b]} U d\mu_\alpha=\overline{\int}_a^b f d\alpha $$ thanks to Lebesgue's monotone convergence theorem.
So far, nothing has been assumed on $f$, except that $f$ is a bounded real function on $[a,b]$. Note that $f \in \mathfrak R(\alpha)$ if and only if its lower and upper integrals coincide, hence if and only if $$\int_{[a,b]} L d\mu_\alpha=\int_{[a,b]} U d\mu_\alpha $$. Since $L \leq U$, this happens if and only if $L(x)=U(x)$ a.e. ($d\mu_\alpha$). In that case $f$ equals to a measurable function a.e. and therefore is measurable itself [because $\mu_\alpha$ is complete?].
Furthermore, if $x$ belongs to no $P_k$, it is quite easy to see that $L(x)=U(x)$ if and only $f$ is continuous at $x$. Since $\cup_k P_k$ is a countable set of continuity points of $\alpha$ it has $\mu_\alpha$-measure zero. So that $L=U$ a.e. if and only if $f$ is continuous a.e.
QED
Best Answer
Hint
These facts will help you develop some idea.
If $f$ or $\alpha$ is discontinuous at countably many points such that the points of discontinuity do not overlap, $f$ is still Riemann-Stieltjes integrable with respect to $\alpha$.
If $f$ and $\alpha$ share a common point of discontinuity and still, $f$ and $\alpha$ are continuous from opposite sites, i.e. left hand limit for $f$ and right hand limit for $\alpha$ are same as their corresponding functional values at those points, or the otherway around (i.e. right hand limit for $f$ and lefthand limit for $\alpha$), $f$ is still Riemann-Stieltjes Integrable with respect to $\alpha$.
If the above condition is not satisfied, the functions are not Riemann-Stieltjes integrable correspondingly.
It will be worth while to attempt to prove the result after you come up with one.