A version of fundamental theorem of calculus

Question

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?

Answer 1

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:

1. $\mathcal{D}_{\delta} \subseteq \Pi$,

2. 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}|} $.

3. 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 $(\dagger)$.