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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.

UPDATE:

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

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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.

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  • $\begingroup$ 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

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