Mathematics Prelims

October 27, 2008

The Lebesgue-Stieltjes Measure

Filed under: Analysis,Measure Theory — cjohnson @ 5:19 pm

Consider an increasing, right-continuous function $F : \mathbb{R} \to \mathbb{R}$.  We can measure the length of an interval $I$ in $\mathbb{R}$ with end points $a$ and $b$ (e.g., ${}[a, b]$) as $\displaystyle \ell_F(I) = F(b) - \lim_{x \to a^-} F(a)$.

(Capinski uses $\ell_F(I) = F(b) - F(a)$ with the interval $I$ restricted to being of the form $(a, b]$, but I believe this gives the same measure in the end.)

Using this definition of the length of an interval, we can then construct an outer measure on $\mathbb{R}$, call it $\mu_F^*$, as follows. $\displaystyle \mu_F^*(A) = \inf\left\{ \sum_{n=1}^\infty \ell_F(I_n) : A \subseteq \bigcup_{n=1}^\infty I_n \right\}$

Where each $I_n$ is a bounded interval.  Proceeding as we would in defining the usual Lebesgue measure on $\mathbb{R}$, we will let $\mu_F$ be a measure on $\mathcal{M}_F$ where $\displaystyle \mathcal{M}_F = \{ E \subseteq \mathbb{R} : \forall A \subseteq \mathbb{R}, \, \mu_F^*(A) = \mu_F^*(A \cap E) + \mu_F^*(A \cap E^\complement) \}$

Now we’ve gone from an increasing, right-continuous function to a measure on $\mathbb{R}$.  Note that sets that were null with the Lebesgue measure, may not be anymore, depending on our choice of $F$.  For instance, if we have $\displaystyle F(X) = \left\{ \begin{array}{ll} x & : x < 0 \\ 1 + x &: x \geq 0 \end{array}\right.$

Then $\mu_F(\{0\}) = 1$, though with the standard Lebesgue measure we have $\mu(\{0\}) = 0$.

It will be convenient to have the convention that if $F$ is an increasing, right-continuous function that $\displaystyle \int_E g \, dF$

is actually short-hand for the Lebesgue integral of $g$ over $E$ using the measure obtained from $F$ as we’ve described above.  This is normally referred to as the Lebesgue-Stieltjes integral with integrator $F$.

1. […] that is an increasing, right-continuous function with and .  Using the ideas from the Lebesgue-Stieltjes measure article, we can have that gives us a measure and sigma-algebra on .  Let and be the measure and […]

Pingback by Distribution of a Random Variable « Mathematics Prelims — November 1, 2008 @ 12:27 pm

2. The first line seems wrong.

The function is defined as right continuous. x tends to a+ not a- .

Comment by N. Srinivasan — April 13, 2010 @ 9:08 am

• F(a+)=F(a) by right continuity, and so, we do not need F(a+); but we need F(a-), since F(b)-F(a) is the measure of (a,b] rather than [a,b]; in fact, F(a)-F(a-) is the measure of {a}=[a,a].

Comment by Boris Tsirelson — February 29, 2012 @ 8:46 am

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