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Let $f: [-1, 1] \rightarrow \mathbb{R}$ and $f(x) = 2x+3$. Also define

\begin{equation*} \alpha(x) = \begin{cases} \hfill 1 \hfill & \text{ if $x = 0$} \\ \hfill 0 \hfill & \text{ if $x \neq 0$} \\ \end{cases} \end{equation*}

Show $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[-1,1]$ and determine the value of $\int_{-1}^{1} fd\alpha(x)$.

My thoughts:

Look at the Riemann-Stieltjes sum over an arbitrary partition $P$, given by $S(P,f,\alpha) = \sum_{P} f(t_k)*(\alpha(x_k)-\alpha(x_{k-1}))$. Notice that $\alpha(x_k) = 0$ for all $k$ unless $x_k = 0$ for some $k$. This means that if our partition $P$ does not contain the point $x = 0$ then the sum $S(P,f,\alpha) = 0$. So assume $P$ contains $x = 0$. Then all but two of terms equal zero and we have \begin{align*} S(P,f,\alpha) &= f(t')*(\alpha(0)-\alpha(x_{k-1})) + f(t'')*(\alpha(x_{x+1})-\alpha(0)) \\ &= f(t') - f(t'') \\ &= 2(t' - t'') \end{align*} where $t' < 0$ and $t'' > 0$.

I also considered using the fact that $f$ is strictly increasing and $\text{min}_{-1 \leq x \leq 1} f(x) = 1$, $\text{max}_{-1 \leq x \leq 1} f(x) = 5$ which means $t'-t''$ is in the worse case $(1-5)$ so $|S(P,f,\alpha)| \leq |2*(1-5)| = 8$.

We want to find a value (the integral), $I \in \mathbb{R}$ such that $\forall \epsilon$ and $\forall P$ which refine some $P_\epsilon$ we have: $|S(P,f,\alpha)-I| < \epsilon$ for all choice of partition tags, $\{t_k\}$. This will also prove $f$ is R-S integrable.

I feel like we can consider a partition so fine that $t'$ and $t''$ are arbitrarily close to $0$ so that our integral is zero, but I don't know whether that's true or how to show that.

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  • $\begingroup$ What definition are you using for RS-integrable function? $\endgroup$ – Juniven Mar 11 '17 at 8:03
  • $\begingroup$ It seems that your problem is a direct consequence of the following THEOREM. If $f$ is continuous on $[a,b]$ and $\alpha$ is a function of bounded variation on $[a,b]$ then the integral $$\int_a^bfd\alpha$$ exists. This result is standard in Stieltjes-type integrals. For instance, see the book of Wheeden, entitled "Measure and Integral: An Intro to Real Analysis", p.27. $\endgroup$ – Juniven Mar 11 '17 at 8:16
  • $\begingroup$ To prove it directly, just choose $$P_\epsilon = \left\{-1, -\frac \epsilon 4, \frac \epsilon 4, 1\right\}$$ Any refinement $P$ of this partition will have $|S(P,f,\alpha)| \le \epsilon$ $\endgroup$ – Paul Sinclair Mar 11 '17 at 16:14

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