Let $\alpha: [a,b]\to\mathbb{R}$ be constant in the intervals $[a,c)$ and $(c,d]$, where $a<c<b$. Show that if $f:[a,b]\to\mathbb{R}$ is continuous in the point $c$, then $\int_a^b f d\alpha = f(c)\cdot [a(c+)-a(c-)]$. However, if $f$ and $\alpha$ are discontinuous in $c$, then there's no integral $\int_a^b f d\alpha$. In particular, given $f:[a,b]\to\mathbb{R}$, if the integral $\int_a^b fd\alpha$ exists for all functions $\alpha:[a,b]\to\mathbb{R}$ of bounded variation, then $f$ is continuous.

Recalling from Stieltjes integration,

$$\int_a^b f(t) d\alpha = \lim_{|P|\to 0} \sum_{}^k f(\zeta_i)[\alpha(t_i)-\alpha(t_{i-1})]$$

  • $\alpha:[a,b]\to\mathbb{R}$ is of bounded variation when $\alpha$ is a rectifiable path in $\mathbb{R}$ (that is, the amount it 'walks' is always finite).

There's a theorem that says: $f$ continuous and $\alpha$ has bounded variation, then $\int_a^b f(t) d\alpha$ exists. We at least know that in the continuous case our integral exists.

Another theorem says that

$f$ continuous and $\alpha$ of class $C^1$ in $[a,b]$, then $\int_a^b f(t) d\alpha = \int_a^b f(t)\alpha'(t) \ dt$

I ca'nt see how to arrive at $f(c)\cdot [a(c+)-a(c-)]$. Specially, because $a\ (t) = 1$ so $\alpha$ disappears in the integration. I don't think this theorem is useful.


I tried:

$$\int_a^b f(t) d\alpha = \int_a^{c-\epsilon} f(t) d\alpha + \int_{c-\epsilon}^{c+\epsilon} f(t) d\alpha + \int_{c+\epsilon}^b f(t) d\alpha$$.

The first and the last integrals are $0$ but the middle one is not... so we end up with: $$\int_a^b f(t) d\alpha = \int_{c-\epsilon}^{c+\epsilon} f(t) d\alpha = \lim_{|P|\to 0} \sum_i^k f(\zeta_i)[\alpha(t_{i}-t_{i-1})]$$

where $P$ is a partition of $(c-\epsilon, c+\epsilon)$. How do I end this? I know I should take the limit with $\epsilon \to 0$ or something like that, but it doesn't make much sense in the expression I have


You cannot use the theorem that says that if $f$ is continuous and $\alpha$ is of class $C^1$, as in this case $\alpha$ is not even continuous, let alone differentiable. In particular, $\alpha$ is generically discontinuous at $c$.

The way to prove this is to go directly from the definition in terms of Riemann-Stieltjes sums. Roughly, if $t_i$ partitions the interval $[a,b]$, then for almost all $i$ we have that $\alpha(t_{i+1}) - \alpha(t_i) = 0$. The only times this does not occur is when $c \in (t_i, t_{i+1})$ or when $c = t_i$ or $c = t_{i+1}$.Thus you have at most two summands in the Riemann-Stieltjes sum that do not vanish.

You then proceed directly through the definition, and use the continuity of $f$ at $c$ to handle the behavior of the (at most two) surviving summands in the Riemann-Stieltjes sum.

Edited to include more

Let $\{t_i\}$ be a partition of $[a,b]$ with maximum width less than $\delta$. Suppose initially that $c \neq t_i$ for any $i$, and I will let you handle the case when $c = t_i$ separately. As noted above, for almost all $t_i$, we have that $\alpha(t_{i+1}) - \alpha(t_i) = 0$, except for the single $j$ with $c \in (t_j, t_{j+1})$.

For such a partition, the Riemann-Stieltjes sum reduces to a single summand, $$ \sum f(\zeta_i) [ \alpha(t_{i+1}) - \alpha(t_i) ] = f(\zeta_j) [ \alpha(t_{j+1}) - \alpha(t_j)].$$ In this expression, $0 < c - t_{j} < \delta$ and $0 < t_{j+1} - c < \delta$, and $\zeta_j \in (t_{j}, t_{j+1})$. As $\alpha$ is constant in $[a, c)$, it's clear that $\alpha(t_j) = \alpha(c^-)$. Similarly, $\alpha(t_{j+1}) = \alpha(c^+)$.

So for such a partition, the Riemann-Stieltjes sum reduces to $$ \sum f(\zeta_i) [ \alpha(t_{i+1}) - \alpha(t_i) ] = f(\zeta_j) [ \alpha(c^+) - \alpha(c^-)].$$ Notice that the only dependence on the partition is how far $\zeta_j$ is from $c$ (and our assumption that $c$ is not a partition point). As one chooses finer partitions, $\zeta_j \to c$ and, since $f$ is continuous at $c$, $f(\zeta_j) \to f(c)$.

Thus the limit as the partition becomes finer exists, and is equal to $$ f(c) [\alpha(c^+) - \alpha(c^-)],$$ as expected.

It remains to you to handle the case when a partition includes $c$ as a partition point. But the work is very similar, and it ultimately reduces to a sum of two summands instead of one.

  • $\begingroup$ Would you mind taking a look at my update? $\endgroup$ – Guerlando OCs Aug 26 '17 at 16:41
  • $\begingroup$ Sure. I've updated my answer to get you further along. $\endgroup$ – davidlowryduda Aug 27 '17 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.