I have three simple questions. I'm working with a problem in an old qualification exam, which asks me to express $\mu(E)$ explicitly, within the settings.
Settings. Let $g:[0,1]\to \mathbb R$ be be a monotonically increasing function. Then we may regard $g$ as a positive regular Borel measure on $[0,1]$ as follows.
Define $\Lambda:C([0,1])\to \mathbb R$ by the Riemann-Stieltjes integral
$$ \Lambda f = \int_0^1 f dg. $$
Then $\Lambda$ is a bounded linear functional of norm $\le g(1)-g(0)$. This can be shown by
$$ \left|\int_0^1 fdg\right| \le \int_0^1 \|f\|_\infty dg = \|f\|_\infty (g(1)-g(0)). $$
So by the Riesz representation theorem on $C_0([0,1])$, we get a unique regular Borel measure $\mu$ such that
$$ \Lambda f = \int_0^1 fd\mu $$
for all $f\in C([0,1])$. In fact, $\mu$ in this case is a positive measure.
Problem. Compute $\mu([a,b])$, for a closed interval $[a,b] \subset [0,1]$.
My answer. Let $E=[a,b]$. Then
$$ \mu(E) = \int_0^1 \chi_E d\mu = \int_0^1 \chi_E dg = g(b)-g(a). $$
The first question. Is my answer correct?
The qual problem does not ask about my second and third question, but I'm just curious about generalizing the problem:
The second question. May I generally say, for a BV function $g$ and a measurable set $E$, that
$$ |\mu|(E) = \text{the total variation of $g$ on the set $E$}? $$
The measure $|\mu|$ denotes of course the total variation measure of $\mu$.
The third question. Then what is $\mu([a,b])$? I guess the answer is
$$ g^{\wedge}(b)-g^\wedge (a) -g^\vee (b) + g^\vee(a),$$
where $g^\wedge$ and $g^\vee$ are the unique monotonically increasing functions such that $g=g^\wedge – g^\vee$, but not sure.
Best Answer
You must pay attention the the question in the exam does NOT assume that $g$ is right (or left)-continuous.
Your first question: Your answer is actually wrong, exactly because it is not assumed that $g$ is right (or left)-continuous.
Counter-example: Given any $c \in [\frac{1}{4}, \frac{3}{4}]$. Let $g_c$ be defined by $g_c(x) = \frac{1}{2}x$ if $x\in [0, \frac{1}{2})$, $g_c(x) = \frac{1}{2}x + \frac{1}{2}$ if $x\in ( \frac{1}{2}, 1]$, and $g_c(\frac{1}{2})) =c$.
Building $\mu$ as described in the qual exam, all those $g_c$ will produce the same $\mu$. So, it is clear that it is not possible that $\mu([a,b]) = g_c(b)-g_c(a)$, because then we would have $$\mu \left(\left[0, \frac{1}{2} \right] \right) = g_c\left( \frac{1}{2} \right) - g_c(0) = c$$ which doe not make sense because $\mu$ does not depend on $c$.
Now, which step in your answer fails?
The equality $\int_0^1 f d\mu = \int_0^1 f dg$ holds only for $f\in C([0,1])$ and $\chi_E \notin C([0,1])$.
So what is the correct answer? Approximate $\chi_E$ from above by continuous functions and we get $$ \mu([a,b]) = \lim_{x\to b^+} g(x) - \lim_{x\to a^-} g(x)$$
Note that, since $g$ is a monotonically increasing function, the lateral limits exists.
Remark: It is also interesting to remark that, we can also prove that $$ \mu((a,b]) = \lim_{x\to b^+} g(x) - \lim_{x\to a^+} g(x)$$
We can define $G(x)= \lim_{y\to x^+} g(y)$. We can prove that $G$ is a right-continuous monotonically increasing function and, using the construction in qual exam, that $G$ and $g$ produces the same $\mu$. So, we can actually replace $g$ by $G$. In this case, we get: $$ \mu((a,b]) = G(b) - G(a)$$ and $$ \mu([a,b]) = G(b) - \lim_{x\to a^-} G(x)$$
Your second question: Yes, that is correct. Sketch of the proof: Since $g$ a BV function, let $g^\wedge$ and $g^\vee$ are the unique monotonically increasing functions such that $g=g^\wedge - g^\vee$. From $g^\wedge$ and $g^\vee$ construct the right-continuous monotonically increasing function $H$ and $K$. Show that $H$ and $K$ are BV function. Define $G=H-K$. Note that $G$ is BV function. The construction in qual exam works for BV functions and $g$ and $G$ will produce same $\mu$. Moreover, given any measurable set $E$, the the total variation of $G$ on $E$ (denoted $TV(G,E)$) is equal to the total variation of $g$ on $E$ (denoted $TV(g,E)$). Then, from a well known result in Measure Theory, we know that $|\mu|(E) = TV(G,E)$, so we get $|\mu|(E) = TV(G,E)= TV(g,E)$.
Your third question: Even if $g$ is BV, the result is NOT correct. See counter-example in answer to the first question.
In a similar way to the first question, we can prove that, if $g$ is a BV function,
$$\mu([a,b]) = \lim_{x\to b^+} g^{\wedge}(x)-\lim_{x\to a^-}g^\wedge (x) - \lim_{x\to b^+} g^\vee (x) + \lim_{x\to a^-}g^\vee(x),$$ where $g^\wedge$ and $g^\vee$ are the unique monotonically increasing functions such that $g=g^\wedge - g^\vee$.