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$Definition: I define the notion of an "increasing partition" $P$ of $[a,b]$ to be a finite partition $\set{z_k}_{k=0}^n \subseteq [a,b]$ where $a = z_0 < z_1 < \cdots < z_n = b$. It's a minor thing that helps reduce the repetition in the wording herein.
Problem Statement: Let $\vp : [a,b] \to \Bbb R$ be a step function with $\set{\a_i}_{i=1}^m$ its discontinuities forming an increasing partition of $[a,b]$. Thus $\vp$ is constant on each interval $(\alpha_{i-1},\alpha_i)$. Let $f : [a,b] \to \Bbb R$ be continuous, and let
\begin{alignat*}{99}
\vp(\alpha_i^+) &:=\;&& \lim_{x \to \a_i^+} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
\vp(\a_i^-) &:=\;&& \lim_{x \to \alpha_i^-} \vp(x) &&\quad \forall i \in \set{0,1,\cdots,m-1} \\
d_i &:=\;&& \vp(\a_i^+) – \vp(\a_i^-) &&\quad \forall i \in \set{1,2,\cdots,m-1} \\
d_0 &:=\;&& \vp(\a_0^+) – \vp(\a_0) \\
d_m &:=\;&& \vp(\a_m) – \vp(\a_m^-)
\end{alignat*}
Show
$$
\int_a^b f \, \dd \vp = \sum_{i=1}^m f(\a_i) d_i
$$
Context: This ultimately comes up as a homework assignment for me, and I just wanted to check my solution, because I have some doubts on it. I'll express those at the end, though, since they tie more intimately into my proof.
Attempted Proof:
Let the following be:
- Let $\g_\a := \set{\a_i}_{i=0}^m$ be an increasing partition of $[a,b]$. (This is the set of discontinuities of $\vp$, i.e. $\vp$ is constant on $(\a_{i-1},\a_i)$ $\forall i$.)
- Let the definitions in the problem statement hold.
- Let $\ve > 0$.
- Let $\ds \delta := \min_{1 \le i \le m} \left| \a_i – \a_{i-1} \right|$.
- Let $\g := \set{x_i}_{i=0}^n$ be an increasing partition of $[a,b]$ with $\norm \g < \delta$.
- Let $\ol \g := \g \cup \g_\a$, and let it be written $\ol \g := \set{\ol{x_i}}_{i=0}^p$.
- Let $\ol{\xi_i} \in [\ol{x_i}, \ol{x_{i-1}}]$ for each $i$. (These are our representative valuation points in the definition of the integral.)
By the merit of $\ol \g$ being a refinement of $\g$, and by merit of definition, we have
$$
\norm{ \ol \g } \le \norm \g < \delta \le \norm{\g_\a}
$$
By the Cauchy criterion for Riemann-Stieljes integration, then, where $R_\g$ denotes the usual Riemann-Stieltjes sum,
$$
\abs{ R_\g – R_{\ol \g} } < \ve
$$
and so $R_{\ol \g}$ equals the desired integral. It remains to see if this equals the desired sum.
Write
$$
R_{\og} = \sum_{j=1}^p f \para{ \ol{ \xi_j } } \Big( \vp \para{ \ol{x_j} } – \vp \para{ \ol{x_{j-1}} } \Big)
$$
From definition, we may have either $\ol{x_i} = \alpha_j \in \g_\a$ for some $i$ (i.e. is a discontinuity of $\vp$), or $\ol{x_i} = x_i \in \g$ for some $i$. Note that for any consecutive pairing $\ol{x_{i-1}},\ol{x_{i}}$ that not both are in $\g_\a$. (This is because each are in $\og$ for which $\norm \og < \delta$, and thus the smallest distance between any two discontinuities of $\vp$ is bigger than the distance between any two members of $\og$.) Thus, we only have three cases to consider, based on how many of a given pair are, indeed, discontinuities:
- Case 1 (neither): In the case neither $\ol{x_{i-1}},\ol{x_{i}}$ are discontinuities of $\vp$, then trivially we will have
$$
\vp \para{ \ol{x_j} } – \vp \para{ \ol{x_{j-1}} } = 0
$$ - Case 2 (right discontinuity): Suppose $\ol{x_{i}} = \a_j$ for some $j$ (but, again, that means $\ol{x_{i-1}}$ is not a discontinuity). Then in such a case, using the piecewise continuity and constantness of $\vp$,
$$
\vp \para{ \ol{x_j} } – \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j } – \vp \para{ \a_{j}^- }
$$ - Case 3 (left discontinuity): Similar to the previous, we have $\ol{x_{i-1}} = \a_j$ for some $j$, and take
$$
\vp \para{ \ol{x_j} } – \vp \para{ \ol{x_{j-1}} } = \vp\para{ \a_j^+ } – \vp \para{ \a_{j} }
$$
Let $\ol{\xi_{i,j}}$ be the $\ol{\xi_i}$ whose terms are not eliminated by merit of Case 1 but lie in Case 2; similarly, $\ol{\xi'_{i,j}}$ for those whose terms lie in Case 3. Then we have, once the terms have negated,
$$
R_{\og} = \underbrace{\sum_{j=1}^m f \para{ \ol{\xi_{i,j}} } \Big( \vp\para{ \a_j } – \vp \para{ \a_{j}^- } \Big) }_{\text{case 2 terms}} + \underbrace{\sum_{j=1}^m f \para{ \ol{\xi'_{i,j}} } \Big( \vp\para{ \a_j^+ } – \vp \para{ \a_{j} } \Big)}_{\text{case 3 terms}}
$$
Note that, for the Case 2 terms, each $\ol{\xi_{i,j}}$ is within $\delta$ of $\a_j$; specifically, $\ol{\xi_{i,j}} \in (\a_j – \delta,\a_j)$. Likewise, $\ol{\xi'_{i,j}} \in (\a_j,\a_j+\delta)$. Since $f$ is continuous, we may take each to be its corresponding $\a_j$, and write
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j } – \vp \para{ \a_{j}^- } \Big) +\sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } – \vp \para{ \a_{j} } \Big)
$$
Some clear cancellation results, giving
$$
R_{\og} = \sum_{j=1}^m f \para{ \a_j } \Big( \vp\para{ \a_j^+ } – \vp \para{ \a_{j}^- } \Big)
$$
Taking the usual convention for what occurs for the $\a_j$ which are $a,b$, we see the $\vp$ difference here is indeed $d_i$. So we see
$$
\int_a^b f \, \dd \vp = R_{\og} = \sum_{j=1}^m f \para{ \a_j } d_i
$$
as desired.
My Questions / Issues:
While I feel like this proof is on the right path, I have some anxieties about some steps. Some in particular I feel aren't quite "there" yet. I was wondering if anyone would be able to help me flesh these ideas uot.
-
The use of the Cauchy criterion for Riemann-Stieltjes integrals (stated here) feels far too un-fleshed out, and I doubt it actually lets me conclude that $R_\og$ equals the integral in question. I imagine it only lets me conclude the integral exists at best. I do see that this proof seems to use something similar towards the end of their proof, but their motivation, choice of bound, and the weird $\delta$'s all escape me.
-
The second chunk of the proof (verifying the alternate form for $R_\og$) feels right, but is there any issue there? Something feels sketchy about how I waited until then to define the $\ol{\xi_{i,j}}$ terms to be $\a_j$ in particular, but I don't see how it would be problematic either…
-
I feel like I barely used the continuity of $f$ at all, which means its purpose is lurking elsewhere. Where would that be?
Thanks for the insights and thoughts you guys can give!
Best Answer
To suggest a better strategy and see the continuity of $f$ enter as it should, consider the proof for a single step discontinuity at $\alpha_1 \in (a,b)$. The objective is to show that for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition with $\|\Gamma\| < \delta$, we have $|R_\Gamma - f(\alpha_1)d_1| < \epsilon$.
Since $f$ is uniformly continuous on the compact interval $[a,b]$ , there exists $\delta > 0$ such that $|f(x) - f(y)| < \epsilon/|d_1|$ when $|x-y| < \delta$.
Take any partition $\Gamma$ with points $a = x_0 < x_1 < \ldots < x_n = b$ such that $\|\Gamma\|< \delta$. There exists and index j such that one of the following three cases must hold:
$$\tag{1} \quad x_{j-1} < \alpha_1 < x_j, \quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$ $$\tag{2} \quad x_{j-1} < \alpha_1 = x_j,\quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$ $$\tag{2} \quad x_{j-1} = \alpha_1 < x_j,\quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$
In each case we have for some intermediate point $\xi \in [x_{j-1},x_j]$,
$$\left| R_\Gamma - f(\alpha_1)d_1\right|= \left|f(\xi)(\varphi(x_j) - \varphi(x_{j-1})-f(\alpha_1) (\varphi(\alpha_1^+) - \varphi(\alpha_1^-) \right|\\ = |f(\xi) - f(\alpha_1)|\, |d_1|< \epsilon$$
To generalize the proof for multiple points where the step function is discontinuous, choose $\delta$ such that in addition to controlling the oscillation of $f$ on subintervals of partitions with $\|\Gamma\| < \delta$, we also have no more than one point of discontinuity $\alpha_k$ in a subinterval.