Chain rule for $e^f$, where $f$ has bounded variation Let $f : \mathbb{R} \to \mathbb{R}$ be a function of normalized bounded variation (NBV), meaning that $f$ is of bounded variation, $f$ is right continuous, and $f(x) \to 0$ and $x \to -\infty$. As explained in Section 3.5 of Folland's Real Analysis textbook, there is a unique complex measure $df$ with the property that $df(x_1, x_2] = f(x_2) - f(x_1)$ for all $x_1 < x_2$ in $\mathbb{R}$.
Moreover, in Exercise 34, which accompanies this section, we are asked to prove that for any two NBV functions $f$ and $g$, we have
$$d(fg) = \tfrac{1}{2}(g(x+) + g(x-))df +  \tfrac{1}{2}(f(x+) + f(x-))dg,$$
where $f(x\pm)$ denotes the right and left hand limits of $f$, respectively.

I would like to know whether the chain rule $d(e^f) = \tfrac{1}{2}(e^{f(x+)} + e^{f(x-)})df$, or a similar formula, is valid.

It seems plausible that a nice formula like this holds for a composition of an NBV function with an exponential. After all, $e^f$ has the same discontinuities as $f$. I have tried to prove this formula by integrating against a test function $\varphi$. By the dominated convergence theorem.
$$\int \varphi \tfrac{1}{2}(e^{f(x+)} + e^{f(x-)})df = \lim_{N \to \infty} \sum_{n= 0}^N \int \frac{f^n(x+) + f^n(x-)}{2(n!)} \varphi df.$$
But I am stuck after this. Hints or solutions are greatly appreciated.
 A: Your version is clearly not completely correct since it doesn't have the right point masses $e^{f_+}-e^{f_-}$.
Since the point parts are easily dealt with by hand, we can perhaps focus on the continuous parts of the measures and simply assume that $f$ is continuous. Then it is indeed true that $d\nu=e^f\, d\mu$, with $d\mu=df$, $d\nu=de^f$. First of all, it is easy to see that $\nu\ll\mu$ ($\nu$ is absolutely continuous with respect to $\mu$; cover a $\mu$-null set by intervals of small total measure). Next, the Radon-Nikodym derivative can be computed $\mu$-a.e. as a pointwise derivative
$$
\frac{d\nu}{d\mu} = \lim_{h\to 0} \frac{\nu(x,x+h)}{\mu(x,x+h)} =
\lim_{h\to 0} \frac{e^{f(x+h)}-e^{f(x)}}{f(x+h)-f(x)} .
$$
By writing
$$
e^{f(x+h)}= e^{f(x)}e^{f(x+h)-f(x)} = e^{f(x)}(1+f(x+h)-f(x))+o(f(x+h)-f(x)) ,
$$
we see that the limit equals $e^{f(x)}$, as required.
A: $\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$No reasonable chain rule will hold here in general, if $f$ is allowed to be discontinuous.
Indeed, let $\mu_f:=df$, the Lebesgue--Stieltjes measure corresponding to the right-continuous function $f$ of bounded variation. Similarly defined is $\mu_{e^f}$.
Let $f_-(x):=f(x-)$ for real $x$.

Claim: There is no chain rule such that for some real $t$ and all right-continuous functions $f\colon\R\to\R$ of bounded variation we would have
\begin{equation*}
    d\mu_{e^f}=((1-t)e^f+te^{f-})\,d\mu_f.  \tag{0}\label{0}
\end{equation*}
More specifically, there is no real $t$ such that for all right-continuous functions $f\colon\R\to\R$ of bounded variation we would have
\begin{equation*}
    \mu_{e^f}([0,1])=\int_{[0,1]}((1-t)e^f+te^{f-})\,d\mu_f. \tag{1}\label{1}
\end{equation*}
In particular, none of the following chain rules will hold for all right-continuous functions $f\colon\R\to\R$ of bounded variation:

*

*$d\mu_{e^f}=e^f\,d\mu_f$;

*$d\mu_{e^f}=e^{f-}\,d\mu_f$;

*$d\mu_{e^f}=\frac{e^f+e^{f-}}2\,d\mu_f$.


Indeed, let
\begin{equation*}
    f:=1_{[0,\infty)}+b\,1_{[1,\infty)}
    =1_{[0,1)}+(1+b)\,1_{[1,\infty)},
\end{equation*}
where $b$ is a real number. Then $f$ is a right-continuous function of bounded variation,
\begin{equation*}
    e^f=1_{(-\infty,0)}+e\,1_{[0,1)}+e^{1+b}\,1_{[1,\infty)},
\end{equation*}
\begin{equation*}
    \mu_f=\de_0+b\de_1,\quad \mu_{e^f}=(e-1)\de_0+(e^{1+b}-e)\de_1,
\end{equation*}
where $\de_x$ is the Dirac measure supported on $\{x\}$. Then \eqref{1} becomes the identity
\begin{equation*}
    e-1+e^{1+b}-e=(1-t)e+t+b[(1-t)e^{1+b}+te].
\end{equation*}
Differentiating this identity twice in $b$, we get $1=(1-t)(2+b)$ for all real $b$, which cannot be true for any given real $t$. $\quad\Box$

On a positive note, if $f$ is continuous, then it can be shown quite elementarily that the chain rule
\begin{equation*}
d\mu_{e^f}=e^f\,d\mu_f  
\end{equation*}
holds. One of the ways to prove this is as follows.
Take any real $a$ and any real $\ep>0$. Let
\begin{equation*}
    E_\ep:=\Big\{x\in[a,\infty)\colon\int_a^y(d\mu_{e^f}-e^f\,d\mu_f-2\ep|d\mu_f|)\le0
    \ \forall y\in[a,x]\Big\}; 
\end{equation*}
note that $\int_a^y$ makes sense here, since the function $f$ is continuous and hence the measures $d\mu_f$, $|d\mu_f|$, and $d\mu_{e^f}$ are non-atomic.
Note also that $a\in E_\ep$, so that $E_\ep$ is nonempty. Let
\begin{equation*}
    s:=\sup E_\ep.
\end{equation*}
We want to show that $s=\infty$. To obtain a contradiction, assume the contrary. Then
\begin{equation*}
    s=\max E_\ep\in E_\ep, 
\end{equation*}
again because the measures $d\mu_f$, $|d\mu_f|$, and $d\mu_{e^f}$ are non-atomic and hence the integral in the definition of $E_\ep$ is continuous in $y$.
Since $f$ is continuous, there is some real $h>0$ such that for all $z\in[s,s+h]$ we have
\begin{equation*}
|e^{f(z)}-e^{f(s)}|\le\ep, 
\end{equation*}
and hence for all $t\in[s,s+h]$
\begin{equation*}
    \int_s^{t}e^f\,d\mu_f=(f(t)-f(s))(e^{f(s)}+\theta_1\ep)
\end{equation*}
and (say by the mean-value theorem)
\begin{equation*}
    \int_s^{t}d\mu_{e^f}=e^{f(t)}-e^{f(s)}=(f(t)-f(s))(e^{f(s)}+\theta_2\ep),
\end{equation*}
where $\theta_1$ and $\theta_2$ stand for certain expressions each with values in $[-1,1]$. So, again for all $t\in[s,s+h]$,
\begin{equation*}
    \int_s^t(d\mu_{e^f}-e^f\,d\mu_f-2\ep|d\mu_f|)
    \le2\ep|f(t)-f(s)|-\int_s^t 2\ep|d\mu_f|\le0  
\end{equation*}
and hence
\begin{equation*}
\int_a^t(d\mu_{e^f}-e^f\,d\mu_f-2\ep|d\mu_f|)
=\int_a^s(d\mu_{e^f}-e^f\,d\mu_f-2\ep|d\mu_f|)
+\int_s^t(d\mu_{e^f}-e^f\,d\mu_f-2\ep|d\mu_f|)
\le0+0. 
\end{equation*}
So, $s+h\in E_\ep$, which contradicts the conditions $s=\max E_\ep$ and $h>0$.
Thus, indeed $s=\infty$, for each real $a$ and each real $\ep>0$. So, for all real $a,y$ such that $a\le y$,
\begin{equation*}
    \int_a^y d\mu_{e^f}\le \int_a^y e^f\,d\mu_f
\end{equation*}
and, similarly,
\begin{equation*}
    \int_a^y d\mu_{e^f}\ge \int_a^y e^f\,d\mu_f. 
\end{equation*}
So, the measures $d\mu_{e^f}$ and $e^f\,d\mu_f$ coincide on the semiring of all intervals. Therefore, $d\mu_{e^f}=e^f\,d\mu_f$. $\quad\Box$
