# 2-dimensional Lebesgue-Stieltjes measure (Bivariate Density to Distribution Function)

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ along with a random variable $X:\Omega\rightarrow\mathbb{R}$ such that $(\mathbb{R},\mathcal{B},\mu)$ is an induced probability space where $\mathcal{B}$ is to denote the Borel $\sigma$-algebra and $\mu$ is a pushforward measure such that $\mu=\mathbb{P}\circ X^{-1}$. Hence, $\mathbb{P}(X\in B)=\mu(B)$ for all $B\in\mathcal{B}$.

It then follows that $\mu$ - as a Lebesgue-Stieltjes measure $\mu(a,b]=F(b)-F(a)$ - is linked to a probability distribution function $F$ such that $F$ is monotonically increasing, right-continuous, and bounded to take values in $\,[0,1]$ with $F(-\infty)=0$ and $F(\infty)=1$.

I do not really feel comfortable with this result, but I was told that

$$\text{d}\mu=\mu(x,x+\text{d}x]=F(x+\text{d}x)-F(x)=\text{d}F.$$

Therefore, one can define the Lebesgue-Stieltjes integral of an arbitrary function $g$ as

$$\int_Bg\,\text{d}F=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}$$

where the righthandside is just the conventional Lebesgue integral w.r.t $\mu$.

Assuming $\mu\ll\lambda$ where $\lambda$ is the 1-dim Lebesgue measure, $F$ is differentiable such that $\text{d}F=F^\prime\,\text{d}x$. It thus follows that $$\int_Bg\,F^\prime\,\text{d}x=\int_{B}g\,\text{d}\mu\qquad\forall\quad B\in\mathcal{B}.$$ Moreover, since $\mu\ll\lambda$ the Radon-Nikodym theorem dictates that there exists a density function $f$ such that $$\int_{B}g\,\text{d}\mu=\int_{B}g\,f\,\text{d}\lambda$$

And finally, since $\lambda$ is a Lebesgue-Stieltjes measure with $F$ such that $x\mapsto x$ such that $\text{d}\lambda=\text{d}x$

$$\int_{B}g\,F^\prime\,\text{d}x=\int_{B}g\,f\,\text{d}\lambda=\int_Bg\,f\,\text{d}x$$ Since this statement holds true for an arbitrary function $g$ as well as for every set $B\in\mathcal{B}$ it would seem to follow that $F^\prime=f$ such that the pdf of a continuous random variable $X$ is simply the derivative of the corresponding cdf.

I would like to be able to transfer this line of reasoning to a bivariate case such that where a vector-valued measurable function $\boldsymbol{X}:\Omega\rightarrow\mathbb{R}^2$ gives a random variable that allows for the construction of an induced probability space $(\mathbb{R}^2,\mathcal{B}_2,\mu)$. At this point $\mathcal{B}_2=\mathcal{B}\otimes\mathcal{B}$ is the product $\sigma$-algebra of $\mathcal{B}$ and $\mu$ is a pushforward measure $\mu=\mathbb{P}\circ \boldsymbol{X}^{-1}$.

Starting from

$\mu(a,b]\times(c,d]=F(b,d)-F(a,d)-F(b,c)+F(a,c)$

(which is just my starting point since $\mathbb{P}(a<X\leq b,c<Y\leq d)=F(b,d)-F(a,d)-F(b,c)+F(a,c)$
and not because I would have read anywhere that the 2-dim Lebegue-Stieltjes measure is defined this way)

I would like to see why $$\text{d}\mu=\text{d}F$$

such that I can arrive at a statement to the effect that

$$\int_{B}g\,\dfrac{\partial^2 F}{\partial x\partial y}\,\text{d}x\,\text{dy}=\int_Bg\,f\,\text{d}x\,\text{d}y.$$

Thank you so much for taking the time to read my question. I would appreciate any pointers you might have for me - especially on why $\text{d}\mu=\text{d}F$ should be true in the bivariate case.

Best regards,
Jon

## 2 Answers

It seems to me that you are going in circles.

Let $\mu$ denote some probability measure on $(\mathbb R^2,\mathcal B_2)$.

Then $\mu$ induces a function $F:\mathbb R^2\to\mathbb R$ that is prescribed by: $$(b,d)\mapsto P((-\infty,b]\times(-\infty,d])$$

This function on its turn induces a (unique) measure $\mu_F$ that is characterized by: $$\mu_F((a,b]\times(c,d])=F(b,d)-F(a,d)-F(b,c)+F(a,c)$$ Observe that we could also write this as:$$\mu_F((a,b]\times(c,d])=\mu((a,b]\times(c,d])$$

So actually now we have two measures $\mu$ and $\mu_F$ that coincide on the collection: $$\mathcal A=\{(a,b]\times(c,d]\mid -\infty<a<b<\infty, -\infty<c<d<\infty\}$$

This collection is closed under intersection so $\mu$ and $\mu_F$ will also coincide on the sigma-algebra generated by $\mathcal A$ which is $\mathcal B_2$.

In a way this closes the circle: $\mu=\mu_F$.

The expression $\text{d}\mu=\text{d}F$ is just a notation for $\mu=\mu_F$.

• Thank you so very much for your answer. So $\text{d}\mu=\text{d}F$ is essentially notational justification for writing $$\int_{B}\,g\,\text{d}\mu=\int_B\,g\,\text{d}\mu_F=\int_B\,g\,\text{d}F?$$ I am afraid, however, since the lhs is evaluated as a Lebesgue-integral and the rhs is a Stieltjes-integral and thus an integral w.r.t. a function $F:\mathbb{R}^2\rightarrow[0,1]$ rather than a measure $\mu:\mathcal{B}\rightarrow[0,1]$, I have to admit, that I still have a hard time seeing why the two need to be equal just as, by definition, $F(b,d)=\mu(-\infty,b]\times(-\infty,d]$. – J.Beck Jul 24 '18 at 14:30
• According to my view every integral is based on a measure. We can write $\int gdF$ but actually that is only a notation for $\int gd\mu_F$ where $\mu_F$ is a unique measure generated by $F$. The notation $d\mu=dF$ tells us that we are dealing with the same measure. Also have a look here. – drhab Jul 24 '18 at 14:55
• If I may be a bit tenacious. Presuming $\mu_F$ is absolutely continuous w.r.t. to $\lambda$, then $F$ is a real-valued continuous function and the Lebesgue-Stieltjes integral specializes to a Rieman-Stieltjes integral. Hence, we can write $\text{d}F=F^\prime\,\text{d}x$ and end up with $$\int\,g\,\text{d}\mu=\int\,g\,\text{d}\mu_F=\int\,g\,\text{d}F=\int\,g\,F^\prime\,\text{d}x$$ which is a simple Rieman integral. I fail to see, how it can just be a question of notation to go from $\int g\,\text{d}\mu$ to $\int g\,F^\prime\,\text{d}x$ – J.Beck Jul 24 '18 at 15:05
• For the last equality $\mu_F$ must be absolutely continuous wrt Lebesguemeasure $\lambda$. Then a density $f$ exists with $\int gdF=\int gfd\lambda$ or in other notation $=\int gfdx$. For this the derivative of $F$ does not have to exist. If it does then $g=F'$ can be chosen as density (there are more than one candidates). – drhab Jul 24 '18 at 15:13
• What I basically want to say is that I feel that there must be a proof to show that this is true iff $F(b,d)=\mu(-\infty,b]\times(-\infty,d].$ – J.Beck Jul 24 '18 at 15:13

Thank to the helpful input of drhab, I believe I have pinpointed my problem:

For the sake of argument I confine my attention to the one-dim case. Let g be a continuous function and $\mu\ll\lambda$ (and therefore also $\mu_F\ll\lambda$). It follows that $F$ is continuous and differentiable.

By the Radon-Nikodym theorem there exists (more than one) density function $f$ such that $\text{d}\mu=f\,\text{d}\lambda$ and $\text{d}\mu_F=f\,\text{d}\lambda$. Hence one can write $$\int_Bg\,\text{d}\mu=\int_Bg\,f\,\text{d}\lambda.$$

Moreover, since $F$ is a differentiable function, it is also true that - in a usual calculus sense - $\text{d}F=F^\prime\,\text{d}x$. Just substituting that in yields $$\int_Bg\,\text{d}F=\int_Bg\,F^\prime\,\text{d}x.$$ On their own, I am okay with both of these statements. I am struggling with the fact, that the notational convention above says that they are equal such that since $$\int_Bg\,\text{d}\mu=\int_Bg\,\text{d}F\quad\text{ it follows that }\quad\int_Bg\,f\,\text{d}\lambda=\int_Bg\,F^\prime\,\text{d}x.$$

I suppose it would already help, if I could see a rigorous proof that

$\,\,\,$a) $\quad\text{d}\lambda=\text{d}x$
(I suppose this says that Rieman and Lebesgue integral coincide provided the Rieman integral exists)

$\,\,\,$b) $\quad F^\prime$ is, in fact, a Radon-Nikodym derivative of $\mu_F$ with respect to $\lambda$.

I am afraid, I could show b) only by relying on the notational convention that $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ (where $\text{d}x$ is essentially $\text{d}F$ of the distribution function prescribed by $x\mapsto x$, i.e. the distribution function corresponding to the Lebesgue measure). Then one could that argue that $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=f\quad\text{and}\quad\dfrac{\text{d}F}{\text{d}x}=F^\prime$$ But since $\text{d}\mu_F=\text{d}F$ and $\text{d}\lambda=\text{d}x$ $$\dfrac{\text{d}\mu_F}{\text{d}\lambda}=\dfrac{\text{d}F}{\text{d}x}\quad\text{such that}\quad f=F^\prime$$ However, this does not answer the question why the derivative of a distribution function is a Radon-Nikodym derivative of the measure induced by said distribution function with respect to Lebesgue measure.

Once again, thank you so much.

Best regards,
Jon

PS. Having read the wikipedia article on "Absolute Continuity", the third paragraph in the section on the "Relation between the two notions of absolute continuity" asserts that
If the absolute continuity holds then the Radon–Nikodym derivative of μ is equal almost everywhere to the derivative of F .

This is precisely what I need to see proved in order to resolve my struggle. Wikipedia cites Royden 1988, Problem 12.17(b) on page 303.