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A measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ that is associated with a monotone increasing and right-side continuous function $F$ is called a Lebesgue-Stieltjes measure. But I am wondering, why it is not true that every measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ is a Lebesgue-Stieltjes measure?

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Every Borel measure is a Lebesgue-Stieltjes, provided it is finite on bounded sets. – Prahlad Vaidyanathan Apr 22 '14 at 17:19
If you assume $\lambda$ to take only values in $\mathbb{R}_{\geq 0}$ then you have a bijection of Lebesgue-Stieltjes measures and equivalence classes of associated monotone increasing and right-side continous functions $F$. But I guess you already know that and you're asking for the case of $\overline{{\mathbb{R_{\ge 0}}}}$ – cQQkie Apr 22 '14 at 17:36
up vote 2 down vote accepted

$\mu$ being a lebesgue-stiltjes measure with corresponding function $F$ implies that $$ \mu\left((a,b]\right) = F(b) - F(a) \text{.} $$

Now take the (rather silly) measure $$ \mu(X) = \begin{cases} \infty &\text{if $0 \in X$} \\ 1 &\text{if $0 \notin X$, $1 \in X$} \\ 0 &\text{otherwise.} \end{cases} $$

We'd need to have $F(x) = \infty$ for $x \geq 0$ and $F(x) = 0$ for $x < 0$ to have $\mu\left((a,b]\right) = F(b) - F(a)$ for $a < 0$, $b \geq 0$. But then $$ \mu\left((0,2]\right) = F(2) - F(0) = \infty - \infty $$ which is

  1. meaningless, and
  2. surely not the same as $1$, which is the actual measure of $(0,2]$.

Note that a measure doesn't necessarily need to have infinite point weights (i.e., $x$ for which $\mu(\{x\}) = \infty$ to cause trouble. Here's another measure on $B(\mathbb{R})$ which isn't a lebesgue-stiltjes measure $$ \mu(X) = \sum_{n \in \mathbb{N}, \frac{1}{n} \in X} \frac{1}{n} \text{.} $$ For every $\epsilon > 0$, $\mu\left((0,\epsilon]\right) = \infty$, which again would require $F(x) = \infty$ for $x > 0$, and again that conflicts the requirement that $\mu((a,b]) = F(b) - F(a) < \infty$ for $0 < a \leq b$. Note that this measure $\mu$ is even $\sigma$-finite! You can write $\mathbb{R}$ as the countable union $$ \mathbb{R} = \underbrace{(-\infty,0]}_{=A} \cup \underbrace{(1,\infty)}_{=B} \cup \bigcup_{n \in \mathbb{N}} \underbrace{(\tfrac{1}{n+1},\tfrac{1}{n}]}_{=C_n} $$ and all the sets have finite measure ($\mu(A)=\mu(B) = 0$, $\mu(C_n) = \frac{1}{n}$).

You do have that all finite (i.e., not just $\sigma$-finite, but fully finite) measures on $B(\mathbb{R})$ are lebesgue-stiltjes measures, however. This is important, for example, for probability theory, because it allows you to assume that every random variable on $\mathbb{R}$ has a cumulative distribution function (CDF), which is simply the function $F$.

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What if both $0$ and $1$ are in $X$? – cQQkie Apr 22 '14 at 17:39
@cQQkie Then the measure is $\infty$. Will make it say that explicitly. – fgp Apr 22 '14 at 17:43

If you have a measure $\mu$ in $\mathbb{R}$ that is finite in compact sets then we define the function

$$ F(x) = \begin{cases} \mu((0,x])&\text{if $x > 0$} \\ -\mu((x,0]) &\text{if $x < 0 $} \\ 0 &\text{if $x=0$} \end{cases} $$

Then $F$ is increasing càdlag function and $\mu = \mu_{F}$. This also shows that every measure in $\mathbb{R}$ which are finite in compact sets is regular.

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Minor correction: It's "càdlàg", continue à droite, limite à gauche. – kahen Apr 22 '14 at 19:10

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