A version of fundamental theorem of calculus Let

*

*$\mu$ be a finite measure defined on the Borel subsets of $\mathbb{R}$ such that $\forall t \in \mathbb{R}, \mu\big(\{t\}\big)=0$;

*$F: \mathbb{R} \to \mathbb{R}$ be a continuously differentiable function;

*$a,b\in \mathbb{R}$ be such that $a < b$.


Is it true that
\begin{equation}
\int_{[a,b]} F'\bigg(\mu\big((-\infty,t]\big)\bigg) \operatorname{d}\mu(t) = F\bigg(\mu\big((-\infty,b]\big)\bigg) - F\bigg(\mu\big((-\infty,a]\big)\bigg)?
\end{equation}

It is obvious that the result holds if $\mu$ is absolutely continuous w.r.t the Lebesgue measure, i.e. if there exists a non-negative $\varphi \in L^1$ such that for all Borel subsets $A$ of $\mathbb{R}$ it holds that $$\mu (A) = \int_A \varphi(t) \operatorname{d}t.$$
In fact, in this case, the function
$$
\mathbb{R} \to \mathbb{R}, t\mapsto F\bigg(\int_{-\infty}^t \varphi(s) \operatorname{d}s\bigg)
$$
is an absolutely continuous function, with derivative given a.e. by
$$
\mathbb{R} \to \mathbb{R}, t\mapsto F'\bigg(\int_{-\infty}^t\varphi(s)\operatorname{d}s\bigg) \varphi(t)
$$
and so by the fundamental theorem of calculus for absolutely continuous functions
\begin{align}
\int_{[a,b]} F'\bigg(\mu\big((-\infty,t]\big)\bigg) \operatorname{d}\mu(t)  &= \int_{[a,b]} F'\bigg(\int_{-\infty}^t\varphi(s)\operatorname{d}s\bigg) \varphi(t) \operatorname{d}t \\ &= F\bigg(\int_{-\infty}^b \varphi(t) \operatorname{d}t\bigg) - F\bigg(\int_{-\infty}^a \varphi(t) \operatorname{d}t\bigg) \\&= F\bigg(\mu\big((-\infty,b]\big)\bigg) - F\bigg(\mu\big((-\infty,a]\big)\bigg).
\end{align}
However, what about the general case?
 A: 0. We introduce the following notation:

*

*For a finite Borel measure $\mu$ on $\mathbb{R}$, we write
$$G(x) = \mu((-\infty, x]).$$


*If $f$ is a function on $[a, b]$, then
$$ \bigl[ f \bigr]_{a}^{b} = f(b) - f(a). $$


*If $f$ is right-continuous and has left-limit at $a$, then we write the jump size of $f$ at $a$ by
$$ f(a^-) = \lim_{x\uparrow a} f(x) \qquad\text{and}\qquad \Delta f(a) = f(a) - f(a^-). $$

1. Suppose that $\mu$ is atomless. Then we indeed have
$$ \int_{[a,b]} (F'\circ G) \, \mathrm{d}\mu = \bigl[ F\circ G \bigr]_{a}^{b}. \tag{$\diamond$} $$
In order to prove this, let $\Pi = \{a = x_0 < x_1 < \dots < x_n = b\}$ be a partition of $[a, b]$. Then
\begin{align*}
\left| \int_{[a,b]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{a}^{b} \right|
\leq \sum_{i=1}^{n} \left| \int_{(x_{i-1}, x_i]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{x_{i-1}}^{x_i} \right|. \tag{1}
\end{align*}
By the mean value theorem, there exist $y^*_i, z^*_i \in [G(x_{i-1}), G(x_i)]$ such that
$$
\int_{(x_{i-1}, x_i]} (F'\circ G) \, \mathrm{d}\mu
= F'(y^*_i) \mu((x_{i-1}, x_i])
= F'(y^*_i) \bigl[ G \bigr]_{x_{i-1}}^{x_i} \tag{2}
$$
and
$$
\bigl[ F\circ G \bigr]_{x_{i-1}}^{x_i} = F'(z^*_i) \bigl[ G \bigr]_{x_{i-1}}^{x_i}. \tag{3}
$$
So it follows that
\begin{align*}
\text{(1)} = \sum_{i=1}^{n} \left| F'(y^*_i) - F'(z^*_i) \right| \bigl[ G \bigr]_{x_{i-1}}^{x_i}.
\end{align*}
Letting the mesh size $\|\Pi\| = \max\{ x_i - x_{i-1} : 1 \leq i \leq n\} \to 0$, this bound converges to $0$ by the uniform continuity of $G$ and $F'$.

2. Now we turn to the general case. We claim that, for any $F \in C^1([a,b])$, the following holds:
$$
\begin{aligned}
\int_{(a,b]} (F'\circ G) \, \mathrm{d}\mu
&= \bigl[ F\circ G \bigr]_{a}^{b} - \sum_{x \in (a, b]} \left( \Delta(F\circ G)(x) - F'(G(x)) \Delta G(x) \right).
\end{aligned} \tag{$\dagger$}
$$
Note that the domain of integration is $(a, b]$ and not $[a, b]$. This is merely for convenience, and the version for the integral over $[a, b]$ may be obtained by adding the term $F'(G(a))\Delta G(a)$ to the right-hand side of $(\dagger)$.
In order to prove $(\dagger)$, let $\mathcal{D}$ denote the set of discontinuities of $G$ on $(a, b]$. Since $\mu$ is finite and $\mathcal{D}$ is at most countable, for each $\delta > 0$ we can find a finite subset $\mathcal{D}_{\delta} \subseteq \mathcal{D}$ such that
$$ \mu(\mathcal{D}\setminus\mathcal{D}_{\delta}) < \delta. $$
Now fix $\epsilon > 0$. Then we can find $\delta \in (0, \epsilon)$ such that
$$ \forall x, y \in [a, b] \ : \quad |x - y| < \delta \ \Rightarrow \ |F'(x) - F'(y)| < \epsilon. $$
Then by noting that $G$ is either continuous or has jump size less than $\delta$ at each point of $(a, b]\setminus\mathcal{D}_{\delta}$, we can find a partition $\Pi = \{a = x_0 < \dots < x_n = b\}$ of $[a, b]$ satisfying the following conditions:

*

*$\mathcal{D}_{\delta} \subseteq \Pi$,


*If $x_i \in \mathcal{D}_{\delta}$, then $(G(x_i^-) - G(x_{i-1})) < \frac{\epsilon}{|\mathcal{D}_{\delta}|} $ and $\left| F(G(x_i^-)) - F(G(x_{i-1})) \right| < \frac{\epsilon}{|\mathcal{D}_{\delta}|} $.


*If $x_i \notin \mathcal{D}_{\delta}$, then $ [G]_{x_{i-1}}^{x_i} < \delta $.
Using this, we may write
\begin{align*}
\int_{(a,b]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{a}^{b}
= \sum_{x \in \mathcal{D}_{\delta}} \left( F'(G(x)) \Delta G(x) - \Delta(F\circ G)(x) \right) + \Sigma_{\text{c}} + \Sigma_{\text{d}},
\end{align*}
where
\begin{align*}
\Sigma_{\text{c}}
&:= \sum_{\substack{1 \leq i \leq n \\ x_i \notin \mathcal{D}_{\delta}}} \left( \int_{(x_{i-1},x_i]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{x_{i-1}}^{x_i} \right), \\
\Sigma_{\text{d}}
&:= \sum_{\substack{1 \leq i \leq n \\ x_i \in \mathcal{D}_{\delta}}} \left( \int_{(x_{i-1},x_i)} (F'\circ G) \, \mathrm{d}\mu - [F(G(x_i^-)) - F(G(x_{i-1}))] \right).
\end{align*}
We study $\Sigma_{\text{c}}$ and $\Sigma_{\text{d}}$ separately.

*

*For each $i \in \{1,\dots,n\}$ such that $x_i \notin \mathcal{D}_{\delta}$, Mean Value Theorem allows to choose $y^*_i$ and $z^*_i$ in $[G(x_{i-1}), G(x_i)]$ satisfying $\text{(2)}$ and $\text{(3)}$. Moreover, $ [G]_{x_{i-1}}^{x_i} < \delta $ implies $\left| y^*_i - z^*_i \right| < \delta$. So
\begin{align*}
\left| \int_{(x_{i-1},x_i]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{x_{i-1}}^{x_i} \right|
\leq \left| F'(y^*_i) - F'(z^*_i) \right| \bigl[ G \bigr]_{x_{i-1}}^{x_i}
< \epsilon \bigl[ G \bigr]_{x_{i-1}}^{x_i}.
\end{align*}
Summing this over all $i \in \{1,\dots,n\}$ satisfying $x_i \notin \mathcal{D}_{\delta}$, we obtain
$$ \left| \Sigma_{\text{c}} \right| \leq \epsilon [G]_{a}^{b}. $$


*For each $i \in \{1,\dots,n\}$ with $x_i \in \mathcal{D}_{\delta}$,
$$ \left| \int_{(x_{i-1},x_i)} (F'\circ G) \, \mathrm{d}\mu \right|
\leq \left( \sup |F'| \right) (G(x_i^-) - G(x_{i-1}))
< \left( \sup |F'| \right) \frac{\epsilon}{|\mathcal{D}_{\delta}|}. $$
So it follows that
$$ \left| \Sigma_{\text{d}} \right| \leq \epsilon \left( 1 + \sup |F'| \right). $$
Together these prove that
$$ \left| \int_{(a,b]} (F'\circ G) \, \mathrm{d}\mu - \bigl[ F\circ G \bigr]_{a}^{b} - \sum_{x \in \mathcal{D}_{\delta}} \left( F'(G(x)) \Delta G(x) - \Delta(F\circ G)(x) \right) \right| < \epsilon \left( 1 + \sup |F'| + [G]_{a}^{b} \right) $$
and letting $\epsilon \to 0^+$ proves $(\dagger)$.
