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Problem

Suppose $f$ is continuous and $\phi$ is of bounded variation on $[a, b]$.

Show that the function $\psi(x)=\int_a^xfd\phi$ is of bounded variation on $[a, b]$.

If $g$ is continous on $[a, b]$, show that $\int_a^b gd\psi=\int_a^b gfd\phi$.

My proof

  1. Since $f$ is continuous $f$ is uniformly continuous on $[a, b]$. That is, $$\forall\varepsilon>0, \exists\delta\gt0 ~s.t.~ \left|f(x)-f(y)\right|\lt\varepsilon~~if~~\left| x-y\right| \lt \delta~~for~~any~~x, y\in[a, b]. $$
  2. Since $\phi$ is of bounded variation on $[a, b]$, we have, for constant $M_1$, $$ S_\Gamma[\phi; a, b]=\sum_{i=1}^m{|\phi(x_i)-\phi(x_{i-1})|} ~~~\le~~~ V[\phi; a, b]=\sup_\Gamma{S_\Gamma} ~~~\le~~~ M_1 ~~~\lt~~~ +\infty $$ for any partition $\Gamma=\left\{a=x_0\lt x_1 \lt \cdots \lt x_m=b\right\}$.
  3. Since $f$ is continuous, $f$ is bounded on $[a, b]$. That is, for constant $M_2$, $$\left| f \right| \le M_2 \lt +\infty$$
  4. $\cdots$
  5. $\cdots$

Is it okay if I considered that $f$ is uniformly continuous and bounded on a specific interval, when given the condition that $f$ is continuous?

Danny_Kim
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  • Why wouldn't it be OK? – Eric Wofsey Mar 30 '16 at 01:48
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    A function continuous on an interval $[a,b]$ IS uniformly continuous, this a good theorem...in fact its true for any compact domain for continuous $f$..Also a continuous function on an interval $[a,b]$ is also bounded for its range is compact, hence bounded. So both assumptions are freebies, really. – Nap D. Lover Mar 30 '16 at 01:52
  • @EricWofsey Since I do not have experience of proving this propositions, I am afraid that I make some jump (non-logical) mistakes. – Danny_Kim Mar 30 '16 at 01:53
  • @LoveTooNap29 Can you let me know the name of theorem, like Heineborel theorem? Moreover, is it also possible that I considered $f$ is bounded? – Danny_Kim Mar 30 '16 at 01:54
  • I edited my comment to address the bounded assumption as well. I've seen the theorem called something in Zorich's book, like Cantor-Cauchy's continuity theorem or something...not a name that stuck. Here's a ref: https://math.stackexchange.com/questions/110573/continuous-function-on-a-compact-metric-space-is-uniformly-continuous.

    Also, more generally if $f:D\to \mathbb{R}$ is continuous, and $D$ a compact space, then $f(D)$ is a compact subset of $\mathbb{R}$, hence closed and bounded by Heine-borel.

     – Nap D. Lover Mar 30 '16 at 01:56
  • Ahhhhh. I do not progress anymore. I failed to prove this. So sad. :( – Danny_Kim Mar 30 '16 at 04:25

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