Edit. To clear up the confusion that I caused, I will define a signed measure here. The literature sometimes calls it "extended signed measure":

Definition. A signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ is a function $$\mu:\text{Borel sets}\to\mathbb R\cup\{\infty\}$$ or a function $$\mu:\text{Borel sets}\to\mathbb R\cup\{-\infty\}$$ such that

  1. $\mu(\emptyset)=0$,
  2. for any disjoint Borel sets $A_1, A_2, A_3, \dots$, we have $$\mu\left(\bigcup_{n\in\mathbb N} A_n\right)=\sum_{n\in\mathbb N} \mu(A_n),$$ with the convention that $\infty+\text{anything}=\infty$ and $-\infty+\text{anything}=-\infty$. Note that $\infty-\infty$ can never occur, since $\{-\infty, \infty\}\subset\operatorname{Image}\mu$ is impossible by definition.

Back to the question. Let $f\in L^1_{\text{loc}}(\mathbb R)$, i.e. $f$ is a locally absolutely integrable function. It is well-known that the distributional derivative of $f$ doesn't have to be expressible as a $L_{\text{loc}}^1$ function again. For example, if $f$ is the characteristic function of $[0,\infty[$ (or the characteristic function of $]0,\infty[$, for that matter), then its distributional derivative corresponds to the Dirac measure $\delta_0$, which has no $L^1_{\text{loc}}$-density with respect to the Lebesgue measure.

Similarly, if we have a measure on $\mathbb R$, its distributional derivative need not be a measure again. Continuing the above example, the distributional derivative of $\delta_0$ is given by the bounded linear operator

\begin{split}\delta_0': \mathcal C_{\text c}^\infty(\mathbb R) &\to \mathbb R \\ \phi&\mapsto-\phi'(0),\end{split}

which is not expressible as a measure on $\mathbb R$.

My question: Does every $L_{\text{loc}}^1$-function have a distributional derivative that can be expressed as a signed measure? More explicitly, if $f\in L_{\text{loc}}^1(\mathbb R)$, does there always exist a signed measure $\mu$ on $(\mathbb R, \text{Borel sets})$ such that

\begin{equation}\tag{*}\label{*}\bbox[15px,border:1px groove navy]{\int_{\mathbb R}\phi\,\mathrm d\mu = -\int_{\mathbb R}\phi'(t)\cdot f(t)\,\mathrm dt}\end{equation}

for every $\phi\in\mathcal C_{\text c}^\infty(\mathbb R)$ ? Note: In particular, I demand that $\int_{\mathbb R}\phi\,\mathrm d\mu$ is well-defined for every $\mathcal C_{\text c}^\infty(\mathbb R)$ (which, since $\mu$ is signed, can be actually quite a messy affair.)


3 Answers 3


If we understand signed Borel measure as a signed measure on the Borel sets such that $|\mu|(K)<\infty$ for every compact $K$, then there is actually a nice characterization all functions which have signed Borel as distributional derivatives:

First, every signed measure is a difference of two positive measures, so we may as well ask which functions have a positive measure as distributional derivative.

If $\mu$ is a positive Borel measure on $\mathbb R$, then $$ f\colon \mathbb R\to\mathbb R,\,f(x)=\begin{cases}-\mu([x,a))&\text{if }x<a\\ \mu((a,x])&\text{if }x\geq a\end{cases} $$ is locally integrable and it is not hard to check that it has weak derivative $\mu$. Since distributional derivatives are unique up to an additive constant, it follows that every $f\in L^1_{\mathrm{loc}}(\mathbb R)$ with $f'=\mu$ has an increasing representative. Conversely, if $f$ has an increasing representative, then $\langle f',\phi\rangle\geq 0$ for every $\phi\in C_c^\infty(\mathbb R)$ with $\phi\geq 0$. It is well-known that such distributions are represented by positive Borel measures.

Therefore $f\in L^1_{\mathrm{loc}}$ has a signed Borel measure as distributional derivative if and only if it has a representative that can be written as a difference of two increasing functions. Note that differences of monotone functions are exactly functions of locally bounded variation. So another way to phrase this result is to say that $f\in L^1_{\mathrm{loc}}$ has a signed Borel measure as distributional derivative if and only if it has a representative of locally bounded variation.

  • $\begingroup$ Very interesting. I didn't know this! :) $\endgroup$ Aug 15, 2021 at 18:34

Consider \begin{split}f:\mathbb R&\to\mathbb R, \\ x&\mapsto \begin{cases}\sin\left(\frac1x\right)&\text{if }x\in (0,1) \\ 0 & \text{else}\end{cases}.\end{split} We have $f\in L^1(\mathbb R)\cap L^\infty(\mathbb R)$.

Consider the set $$A=\{x\in (0,1), \cos(1/x)>0\}.$$ Then $$\int_A\,\mathrm df=\infty,\qquad \int_{(0,1)\setminus A}\,\mathrm df=-\infty,$$ so $\mathrm df$ is not a signed measure, $\int_{(0,1)} \,\mathrm df$ is not well-defined.

Even assuming that $f$ is continuous doesn't help, see $x(1-x)\sin(1/x^2) 1_{x\in (0,1)}$.

  • 2
    $\begingroup$ I noticed only after asking the question how you run into terrible subtleties with signed measures: For example, the integral is defined through $$\int_{\mathbb R} \phi\,\mathrm d\mu=\int_{P} \phi\,\mathrm d\mu-\int_{N} \phi\,\mathrm d(-\mu),$$ where $(P, N)$ is the Hahn-decomposition of $\mu$. It seems to me that hence, even if $\int_{\mathbb R} \phi\,\mathrm d\mu$ is well-defined for all test functions $\phi$, the integral $\int_{\mathbb R} 1_{[a,b]} \,\mathrm d\mu$ doesn't even have to well-defined! In short, it seems that signed measures have some crazy properties. $\endgroup$ Mar 23, 2021 at 1:36
  • 1
    $\begingroup$ Alternatively the spirit of my answer is that (with $d\mu=df$) $\mu([1/a,1/2])$ is well-defined for $a>2$ but $\lim_{a\to \infty}\mu([1/a,1/2])$ diverges, so $\mu$ is not sigma additive: we can't decompose $(0,1/2)$ as a countably infinite disjoint union $\bigcup U_n$ of Borel sets and get $\mu((0,1/2))=\sum_n \mu(U_n)$ independent of the chosen $U_n$. $\endgroup$
    – reuns
    Mar 23, 2021 at 1:52
  • $\begingroup$ I have cleared up my confusion by editing the question to include a proper definition of a signed measure 🙂. Now, $\infty-\infty$ is avoided so $\int_{\mathbb R} 1_{[a,b]}$ is well-defined in $\mathbb R\cup\{\infty\}$ or $\mathbb R\cup\{-\infty\}$. $\endgroup$ Mar 23, 2021 at 10:05
  • 1
    $\begingroup$ In fact, it turns out that with the proper definition of a signed measure, not even $f(x)=|x|$ has a measure-derivative (even though the heavy-side function is a weak derivative in $L^1_{\text{loc}}$ !). $\endgroup$ Mar 23, 2021 at 10:15
  • $\begingroup$ Ok I have finally looked at your very nice construction. Can I still ask you the following: Suppose that $\mu$ is a measure satisfying (*). Is it then easy to prove that $\mu = \mathrm df$ (which would be some sort of uniqueness result) and how do you prove that $\int_A \,\mathrm df=\infty, \int_{]0,1[\setminus A} \,\mathrm df=-\infty$ ? $\endgroup$ Mar 23, 2021 at 12:44

It seems that the OP is confusing signed measure by Radon measure.

In general, if $f\in L^{loc}_1(\mathbb{R})$, then $\nu^f(dx)=f(x)\,dx$ is not a signed-measure for $\mu(\mathbb{R})$ may be undefined. Example: $f(x)=x$; $g(x)=\sin x$, etc.

When consider as a functional on $C_{00}(\mathbb{R})$ however, $\nu^f$ satisfies the following property:

Property R: For any sequence $\{\phi_n:n\in\mathbb{N}\}\subset\mathbb{C}_{00}(\mathbb{R})$ that is supported an a compact set $K$ (i.e., $\operatorname{supp}(\phi_n)\subset K$ for all $n\in\mathbb{N}$) if $\phi_n\xrightarrow{n\rightarrow\infty}\phi$ uniformly on $\mathbb{R}$, then $$\begin{align}\nu^f(\phi_n)\xrightarrow{n\rightarrow\infty}\nu^f(\phi)\tag{R}\label{R}\end{align}$$

Functionals $\nu$ that satisfy \eqref{R} are called Radon measures A decomposition similar to the Hahn decomposition for signed measures exists for Radon measures or general $\sigma$-continous elementary integrals with finite variation (see Bichteler, K., Integration: A functional approach. Birkhäuser Advanced Texts Basler Lehrbücher. 1998th Edition).

Observation: A regular signed measure is a Radon measure; a Radon measure is not necessarily a signed measure. The restriction of a Radon measure to a compact set $K$ defines a (signed) measure supported in $K$.

Now, if $F$ and $G$ are functions on $\mathbb{R}$ of local bounded variation (i.e. $F$, and $G$ have finite variation on any bounded closed interval $[a,b]$ then the Lebesgue integration by parts formula gives $$\begin{align} \int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\mu_F(dt)\tag{1}\label{one}\end{align}$$ where $\mu_F$ and $\mu_G$ are the Radon measures induced by $F$ and $G$ (i.e. $\mu_F((a,b])=F(b)-F(a)$, and $\mu_G((a,b])=G(b)-G(a)$ for all $-\infty<a\leq b<\infty$). If $\phi=F\in\mathcal{C}^\infty_{00}(\mathbb{R})$, then $\mu_\phi(dx)=\phi'(x)\,dx$ and \eqref{one} takes the form $$\begin{align} \int_{(a,b]}\phi(t)\mu_G(dt)=\phi(b)G(b)-\phi(a)G(a)-\int_{(a,b]}G(t-)\phi'(t)\,dt\tag{2}\label{two} \end{align}$$ If $\operatorname{supp}(\phi)\subset [a,b]$, then $\mu_\phi(dx)=\phi'(x)\,dx$ and \eqref{two} becomes $$\begin{align} \int_\mathbb{R} \phi(t)\,\mu_G(dt)&=\int_{(a,b]}\phi(t)\mu_G(dt)\\ &=-\int_{(a,b]}G(t-)\phi'(t)\,dt=-\int_{\mathbb{R}}G(t-)\phi'(t)\,dt\tag{3}\label{three} \end{align}$$ If $G(t)=\int^t_0 g(s)\,dx$ where $g\in L^{loc}_1(\mathbb{R})$, meaning $G(t)=\int_{[0,t]} g(s)\,dx$ if $t\geq0$ and $-\int_{[t,0]}g(s)\,ds$ when $t<0$, then \eqref{three} takes the form $$\begin{align} \int_{\mathbb{R}}\phi(t)\,\mu_G(dt)=\int_\mathbb{R} \phi(t)\,g(t)\,dt=-\int_{\mathbb{R}}G(t)\phi'(t)\,dt\tag{4}\label{four} \end{align}$$

  • $\begingroup$ Thank you! (Many high quality answers coming in recently :) ) $\endgroup$ Aug 15, 2021 at 21:07

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