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Prove that if $f$ is continuous on $[a,b]$ and $g$ is bounded variation on $[a,b],$ then $$\vert\int_a^bfdg\vert\le [sup_{a\le t \le b} \vert f(t) \vert] V_{[a,b]}g$$

Proof:

As f is continuous on [a,b] and g is BV([a,b]) then f is riemann-stieltjes integrable, i.e. $f\in R(g)$.

But I don't know how to prove $\vert\int_a^bfdg\vert\le [sup_{a\le t \le b} \vert f(t) \vert]$ sup{$\sum_{k=1}^n|g_k(x)-g_{k-1}(x)|: \{x_0,x_1,...,x_n\}\in P[a,b] \}$.

How can I continue the proof?

Can someone give me a hint/solution?

Note: $V_{[a,b]}g$ means the total variation of g on [a,b]

RRL
  • 90,707
user441848
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1 Answers1

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The "straightforward" proof relies on many other results pertaining to the Riemann-Stieltjes integral with respect to a general integrator function of bounded variation.

Let the function $h$ be defined as the total variation of $g$ on the interval $[a,x]$:

$$h(x) = V_{[a,x]}g$$

It can be shown that $f$ is Riemann-Stieltjes integrable with respect to $h$ and $h - g$. Also if $f$ is integrable then so is $|f|$.

Now take $\alpha = (g + h)/2$ and $\beta = (h-g)/2$. Both are increasing functions on $[a,b]$ and using basic properties of Riemann-Stieltjes integration we have

$$\begin{align}\left|\int_a^b f \, dg \right| &= \left|\int_a^b f \, d(\alpha - \beta) \right| \\ &= \left|\int_a^b f \, d\alpha - \int_a^b f \, d\beta \right| \\ &\leqslant \left|\int_a^b f \, d\alpha \right| + \left|\int_a^b f \, d\beta \right| \\ &\leqslant \int_a^b |f| \, d\alpha + \int_a^b |f| \, d\beta \\ & = \int_a^b |f| \, d(\alpha + \beta) \\ &= \int_a^b |f| \, dh \\ &\leqslant \sup_{t \in [a,b]}|f(t)|(h(b) - h(a)) \\ &= \sup_{t \in [a,b]}|f(t)|V_{[a,b]}g\end{align}$$

RRL
  • 90,707