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.



Chain rule for $e^f$, where $f$ has bounded variation
$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$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, $d\mu\_{e^f}=e^f\,d\mu\_f$. $\quad\Box$


