15

Are right-continuous function from $\mathbb{R}^n$ to $\mathbb{R}$ necessarily semi-continuous? If not, are they necessarily Borel measurable? Is there a topological characterization of right-continuous functions (as there is of continuous ones)? Are CDFs of $n$-dimensional random vectors measurable?

Note: A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous iff it is right-continuous at every point $x \in \mathbb{R}^n$. A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous at $x \in \mathbb{R}^n$ iff given any infinite sequence of points in $\mathbb{R}^n$ $(y_0,y_1,\dots)$ that converges to $x$ from above (i.e. the sequence converges to $x$ in the usual, Euclidean sense and in addition every sequence element is greater than or equal to $x$ component-wise), the sequence $(f(y_0), f(y_1), \dots)$ converges to $f(x)$ in the usual sense.

Henry T. Horton
  • 18,984
  • 6
  • 65
  • 80
Evan Aad
  • 11,818
  • 3
    You might want to add the definition of right-continuity for a function defined on $\mathbb R^n$ when $n\gt1$. – Did Aug 02 '12 at 08:28
  • 2
    What are the definition of right-continuous function, and semi-continuous? – Paul Aug 02 '12 at 08:44

2 Answers2

15

Here is a proof that any function $F: \mathbb{R}^n \to \mathbb{R}$ which is right continuous is Borel measurable.

Define $F_n(x) = F\left(\frac{[nx]+1}{n}\right)$, where $[x]$ is the greatest integer smaller than $x$. Clearly $F_n$ are measurable since they are step functions.

From the right continuity of $F$ we get that $F_n\to F$, since $\frac{[nx]+1}{n} \downarrow x$.

Hence since $F$ is the pointwise limit of measurable functions, it is measurable too.

Milind
  • 3,944
  • Very Nice, Milind! – An old man in the sea. Aug 24 '17 at 13:58
  • 1
    to make this a bit more solid, you would need to clarify that $n=1$ or properly define $[x]$ for an $n$-vector $x$. – mathlete Oct 12 '18 at 11:58
  • Let us assume that $x \in \mathbb R$ and $[x]$ is the greatest integer not exceeding $x.$ Then what makes you believe that $\frac {[nx] + 1} {n}$ is decreasing? Take $x = \frac {1} {3}$ and $n = 2,3.$ – Akiro Kurosawa Nov 04 '23 at 06:01
  • Instead you can take a decreasing subsequence of the sequence $\left {\frac {[nx] + 1} {n} \right }_{n \geq 1}$ and argue accordingly. – Akiro Kurosawa Nov 04 '23 at 06:19
  • Another remark what I want to make is that $F_n$ is not a step function as $F_n$ is not a finite linear combinations of the indicators. Rather it is the pointwise limit of a sequence of step functions which is Borel measurable as pointwise limit of a sequence of Borel measurable functions is Borel measurable. – Akiro Kurosawa Nov 04 '23 at 06:45
  • @AkiroKurosawa $\frac {[nx] + 1} {n}$ might not be decreasing, but $\frac {[nx] + 1} {n}>x$, which is enough for the proof (I thought the answer meant "tends to from above" by $\downarrow$). – Asigan Jan 11 '25 at 00:06
  • This is really a proof from the Book! – Asigan Jan 11 '25 at 00:06
11

The answer to the first question is no, even in the case $n=1$: The characteristic function of the half open interval $[0,1)$ is right continuous, but neither upper nor lower semicontinuous.

A right continuous function $\mathbb{R}\to\mathbb{R}$ is indeed Borel measurable. By definition, the inverse image $E$ of an open set has the property that for any $x\in E$, there is some $\delta>0$ so that $[x‚x+\delta)\subseteq E$. It follows that $E$ is a countable union of half open intervals, and hence is Borel measurable. I am not sure about the answer to this one when $n>1$ (the countable union argument no longer holds), but my guess is that right continuous functions are still measurable.

Topological characterization: If we write $\le$ for pointwise comparison on $\mathbb{R}^n$, we can make a one-sided topology by declaring a set $V\subseteq\mathbb{R}^n$ to be open if, for each $x\in V$, there is some $\delta>0$ so that $\{y\ge x\colon\lvert y-x\rvert<\delta\}\subseteq V$. Then the right continuous maps $\mathbb{R}^n\to\mathbb{R}$ are just the ones that are continuous from this topology to the usual topology on $\mathbb{R}$.

I am not too sure on the CDF question either. (I assume CDF stands for cumulative distribution function, in the sense of $F(x)=\mathrm{P}\{X\le x\}$, where $X$ is a random $n$-vector.) It might help that $F$ is not only right continuous, but also monotone. So the set $\{x\colon F(x)\le p\}$ has a particularly simple structure; I imagine it must be measurable, but right now I don't see a proof.

Harald Hanche-Olsen
  • 32,608
  • 2
  • 61
  • 87
  • Thanks, Harald, but it is in fact the case n>1 that i am interested in. See my clarification of the term "right-continuous" in a comment to my original post. Also, of the four questions i posed in my original post i am most interested in the last one concerning CDF. – Evan Aad Aug 02 '12 at 10:49
  • @EvanAad: Okay. I'll let my answer stand, though, since at least it provides a simple, partial answer which may be of use. I even expanded on it, though it is still far from a satisfactory answer. – Harald Hanche-Olsen Aug 02 '12 at 12:33
  • Thank you, Harald. Despite your reservations, i think you did actually answer my question re CDFs (and yes, i meant "CDF" as in "Cumulative Distribution Function"), since the one-sided topology you defined is contained in (in fact, generates) the Borel field on R^n (at least so it appears to me, at a cursory glance). – Evan Aad Aug 02 '12 at 13:13
  • In fact, Harald - and please correct me if i'm wrong - the one-sided topology you described can be generated by sets that are the result of intersecting a "generalized ray" of the form [x, infty) (x in R^n) with an open set in R^n. – Evan Aad Aug 02 '12 at 13:28
  • 1
    @EvanAad: This topology appears to be the product of $n$ copies of the Sorgenfrey line. But I don't quite see why the open sets of this topology are Borel. It certainly has a basis of Borel sets, but I don't think it's second countable, so why does it follow that any union of such sets is also Borel? – Nate Eldredge Aug 02 '12 at 21:54
  • @NateEldredge: Right you are. Thanks for pointing this out. So the question of CDF measurability is still open. – Evan Aad Aug 03 '12 at 05:25
  • 1
    @EvanAad: In fact, let $E$ be a non-measurable subset of $\mathbb{R}$ and let $A = \bigcup_{x \in E} [x,\infty)^2$. $A$ is open in our topology on $\mathbb{R}^2$ (the Sorgenfrey plane), but not Borel with respect to the usual topology, since its intersection with the diagonal line $y=-x$ is a skewed copy of $E$. One might think one could exploit this to make a right-continuous function which is not Borel, but I still don't see how. – Nate Eldredge Aug 03 '12 at 13:11
  • 1
    @NateEldredge: Did you mean $\bigcup_{x\in E}[x,\infty)\times[-x,\infty)$? – Harald Hanche-Olsen Aug 03 '12 at 13:51
  • 1
    @HaraldHanche-Olsen: Oops, yes, I did. – Nate Eldredge Aug 03 '12 at 14:25
  • @NateEldredge: Thanks for your input! I'm pleased to inform the CDF question has been resolved in another thread, so once a counterexample is produced demonstrating a right-continuous function that is not Borel-measurable, i'll be able to mark this thread as "Answered." – Evan Aad Aug 04 '12 at 05:48
  • I'm caving in and marking this thread as "answered", even though the question of whether a $\mathbb{R}^n\rightarrow\mathbb{R}$ right-continuous function can fail to be Borel-measurable has not yet been resolved for $2\leq n$. What i was really after in starting this thread was the part concerning the CDF and this has been resolved in another thread, as mentioned above. – Evan Aad Sep 04 '12 at 07:11
  • Sorry to dig up an old answers but can you please explain why the union is countable? It would seem to me that $\cup_x [x+\delta(x))$ is uncountable. – JohnK Jan 02 '18 at 09:56