"Measure Theory and Probability Theory" by Athreya and Lahiri introduces Lebesgue–Stieltjes measure construction on $\mathbb{R}^n$ in general in the following way:

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ satisfy the following properties:

  1. Let $(x_1,x_2)\leq (y_1,y_2)$ iff $x_1\leq y_1$ and $x_2\leq y_2$. If $x\leq y$ in this partial order, then $f(y_1,y_2)-f(x_1,y_2)-f(y_1,x_2)+f(x_1,x_2)\geq 0$.
  2. $f$ is right-continuous for each coordinate.
  3. $x\leq y\ (\forall x_n\leq y_n)\implies f(x)\leq f(y)$ (this condition is omitted in the book, I think this might be needed).

Then $\mu((x_1,x_2]\times (y_1,y_2])=f(y_1,y_2)-f(x_1,y_2)-f(y_1,x_2)+f(x_1,x_2)$ defines a locally finite Borel measure on $\mathbb{R}^2$. Conversely, any locally finite Borel measure is generated by such a function.

I have some questions about this construction:

  1. I want to see a complete proof of this construction in $\mathbb{R}^n$; the first condition can be defined analogously in $\mathbb{R}^n$. Even in the book, there was no complete proof of this general construction. I am also not sure whether the third condition can be omitted or not.
  2. I wonder if there is a one-to-one correspondence between such functions and locally finite Borel measures on $\mathbb{R}^n$. In the one-dimensional case, a locally finite Borel measure corresponds to a right-continuous increasing function up to a constant. It seems like such a correspondence result, if it exists, should be more restrictive; $f(x,y)=x+y$ yields the zero measure, but any $f(x,y)=F(x)+G(y)$ does so as well.

Any reference or complete construction would be much appreciated.

(The question was originally asked in stackexchange.)

[Edited: Added the authors.]

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    $\begingroup$ You write "$x \le y\ (\forall x_n \le y_n) \implies f(x) \le f(y)$". I think this is supposed to be "$x \le y \implies f(x) \le f(y)$", right, and the parenthetical is just a reminder about the definition of the order? (Also, presumably, $\forall$ is meant to be $\forall n$, right?) I suspect that, if this condition is omitted, it is because the authors are allowing signed measures. $\endgroup$
    – LSpice
    Mar 27, 2021 at 19:26
  • $\begingroup$ This seems to me much better left as an MSE question. \\ The advantage of working with signed measures is that equivalence classes are much nicer; you can reduce your question about when two $f$'s yield the same Lebesgue–Stieltjes measure to questions about when a single $f$ yields the 0 measure. $\endgroup$
    – LSpice
    Mar 27, 2021 at 20:06
  • $\begingroup$ @LSpice It was before when signed measures were introduced. $\endgroup$
    – Taxxi
    Mar 27, 2021 at 22:49
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    $\begingroup$ @LSpice: for signed measures you would be missing a "total bounded variation" hypothesis. $\endgroup$ Mar 29, 2021 at 20:01
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    $\begingroup$ @Taxxi: The measure associated to f does not change if you add to f a function of the form g(x_1)+h(x_2), and this operation does not preserve monotonicity. The condition (1) is a weaker form of monotonicity which seems sufficient in order to define a measure. $\endgroup$ Mar 29, 2021 at 20:06

1 Answer 1


$\newcommand\R{\mathbb R}$Concerning your question 1: A complete proof of this result (based on a discrete approximation) can be found, for instance, in Kallenberg's book "Foundations of Modern Probability", where it is given as Corollary 3.26. Another kind of arguments, using e.g. semirings, can be found e.g. in Example 1.54 in Probability Theory, A Comprehensive Course, by Klenke, where the Lebesgue measure over $\R^d$ is constructed; the construction of the Lebesgue–Stieltjes measure is quite similar.

Concerning your question 2: Clearly, if $\mu_f$ denotes the Lebesgue–Stieltjes measure for a right-continuous function $f\colon\R^d\to\R$ with nonnegative increments (in the natural sense, as defined e.g. in Kallenberg's book), and if $g\colon\R^d\to\R$ is such that \begin{equation} \label{star} \tag{$*$} g(x_1,\dotsc,x_d)=f(x_1,\dotsc,x_d)+\sum_{j=1}^d h_j(x_1,\dotsc,x_{j-1},x_{j+1},\dotsc,x_d) \end{equation} for some functions $h_j\colon\R^{d-1}\to\R$ and all $(x_1,\dotsc,x_d)\in\R^d$, then $\mu_g=\mu_f$. Vice versa, if $\mu_g=\mu_f$, then the (appropriate $d$-fold) increments of $g-f$ are identically zero, which implies \eqref{star}.

  • $\begingroup$ I cannot thank you enough. I have actually put a bounty to the original stackexchange question, so you might want to answer it as well. $\endgroup$
    – Taxxi
    Mar 29, 2021 at 23:36
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    $\begingroup$ @Taxxi : I am glad this was of help. $\endgroup$ Mar 29, 2021 at 23:51

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