# Does every $L^1_{\text{loc}}$-function have a signed measure as a distributional derivative?

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.)

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 $$f\colon \mathbb R\to\mathbb R,\,f(x)=\begin{cases}-\mu([x,a))&\text{if }x is locally integrable and it is not hard to check that it has weak derivative $$\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.

• Very interesting. I didn't know this! :) 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 $$x(1-x)\sin(1/x^2) 1_{x\in (0,1)}$$.

• 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. Mar 23, 2021 at 1:36
• 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 $U_n$. Mar 23, 2021 at 1:52
• 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 $\mathbb R\cup\{-\infty\}$. Mar 23, 2021 at 10:05
• 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 $L^1_{\text{loc}}$ !). Mar 23, 2021 at 10:15
• 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 $\int_A \,\mathrm df=\infty, \int_{]0,1[\setminus A} \,\mathrm df=-\infty$ ? 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 $$-\infty). If $$\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}

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