A question from my homework:

Suppose $\mu$ is a measure on the real line w.r.t to the Borel $\sigma$-algebra such that $\forall x \in \mathbb{R}$ $\mu(\{x\})=0 $. is $\mu$ necessarily absolutely continuous w.r.t. to the Lebesgue measure?

We say $\mu$ is absolutely continuous w.r.t. to $m$ if for every measurable set $E$ such that $m(E)=0$ than $\mu(E)=0$ also.

I can't see a reason why this should be true, but can't find a good counter example. I tired some variations on the counting measure. Also tried to construct some measures that give positive measure on uncountable sets and zero measure on countable, be these come out non-additive (i.e. not measures).

Would greatly appreciate the help.



No, there are non-atomic measures whose support is a set of measure zero.

These are called singular continuous measures and they are not absolutely continuous with respect to the Lebesgue measure.

For example, the measure corresponding to the devil staircase function has as it's support the Cantor set, but the measure of every point is zero.

Let $F$ be the devil staircase function on the unit interval. Then we can define a Borel measure on the unit interval which is singular with respect to the Lebesgue measure and has no atoms as follows: for $E\subset[0,1]$ set

$$\mu_*(E)=\inf\sum_{j=1}^\infty F(b_j)-F(a_j)\tag{1}$$

where the infimum goes over all coverings of $E$ with intervals $(a_j,b_j)$, $j=1,2,\dots$.

This is the Riemann-Stieltjes outer measure associated to $F$. Go ahead and prove that it really is an outer measure. Then we apply Caratheodory's extension theorem (as in the construction of the Lebesgue measure) to obtain a corresponding Borel measure $\mu$ (such that $\mu_*(E)=\mu(E)$ for measurable $E$).

Note that $\mu((a,b))=F(b)-F(a)$. Since $F$ is continuous, it really follows that $\mu(\{x\})=F(x)-F(x)=0$. On the other hand, the measure is not absolutely continuous, because $\mu(E)=\mu(C\cap E)$ for every measurable $E$, where $C$ is the Cantor set. Since $\lambda(C)=0$ ($\lambda$ being the Lebesgue measure) we see that $\mu$ is singular with respect to the Lebesgue measure.

Since you wanted a measure on the real line just extend this measure on the unit interval by $0$ outside of $[0,1]$.

This construction works with any increasing function and it provides (after proper normalization) a 1:1 correspondence between increasing function and positive Borel measures (see Stein, Shakarchi Real Analysis).


Every positive Borel measure $\mu$ (say on the unit interval, it also works on $\mathbb{R}$ but there you need $\sigma$-finiteness as additional assumption) can be uniquely decomposed as


where $\mu_a$ is absolutely continuous, $\mu_{sc}$ is a singular continuous measure (i.e. singular and no atoms) and $\mu_{pp}$ is a pure point measure, i.e. the support consists entirely of atoms (all with respect to Lebesgue measure, but also other measures can be used).

Look up the Radon-Nikodym theorem in Rudin's Real and Complex Analysis for a reference.

  • $\begingroup$ Should it read $\mu(E) = \mu(C \cap E)$ instead of $\mu(C) = \mu(C \cap E)$? $\endgroup$ – gerw Jul 6 '16 at 7:02
  • $\begingroup$ @gerw: Yes, certainly! Edited. Thank you! $\endgroup$ – J.R. Jul 8 '16 at 20:03

No. Every Borel measure which does not contain Dirac measures (non-atomic) and is singular with respect to the Lebesgue measure has this property.

For example the uniform probability measure on the Cantor set which is defined as follows. Let $C$ be the Cantor set which is defined as $C=\cap_{n=1}^\infty I_n$, where $I_n$ is a union of $2^n$ disjoint closed interval, each of length $3^{-n}$. Then define the functional $$ \ell_n(f)=\frac{3^n}{2^n}\int_{I_n}f\,dx, \quad f\in C[0,1]. $$ Clearly, $|\ell_n(f)|\le \|f\|_{\infty}$. It is not hard to show that $\{\ell_n\}$ converges weakly in ${\mathcal M}[0,1]$, and its limit $\ell$ is realized by a positive Borel measure $\nu$, i.e., $$ \ell(f)=\int_{[0,1]}f\,d\nu, $$ which has the required properties.


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