6
$\begingroup$

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.)

$\endgroup$
0

3 Answers 3

Reset to default
7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Very interesting. I didn't know this! :) $\endgroup$ Commented Aug 15, 2021 at 18:34
3
$\begingroup$

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 $x(1-x)\sin(1/x^2) 1_{x\in (0,1)}$.

$\endgroup$
6
  • 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. $\endgroup$ Commented 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 $U_n$. $\endgroup$ Commented 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 $\mathbb R\cup\{-\infty\}$. $\endgroup$ Commented 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 $L^1_{\text{loc}}$ !). $\endgroup$ Commented 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 $\int_A \,\mathrm df=\infty, \int_{]0,1[\setminus A} \,\mathrm df=-\infty$ ? $\endgroup$ Commented Mar 23, 2021 at 12:44
1
$\begingroup$

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 $$\begin{align} \int_{\mathbb{R}}\phi(t)\,\mu_G(dt)=\int_\mathbb{R} \phi(t)\,g(t)\,dt=-\int_{\mathbb{R}}G(t)\phi'(t)\,dt\tag{4}\label{four} \end{align}$$

$\endgroup$
1
  • $\begingroup$ Thank you! (Many high quality answers coming in recently :) ) $\endgroup$ Commented Aug 15, 2021 at 21:07

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.