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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.

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You might want to add the definition of right-continuity for a function defined on $\mathbb R^n$ when $n\gt1$. –  Did Aug 2 '12 at 8:28
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What are the definition of right-continuous function, and semi-continuous? –  Paul Aug 2 '12 at 8:44

2 Answers 2

up vote 3 down vote accepted

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.

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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 2 '12 at 10:49
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@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 2 '12 at 21:54
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@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 3 '12 at 13:11
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@NateEldredge: Did you mean $\bigcup_{x\in E}[x,\infty)\times[-x,\infty)$? –  Harald Hanche-Olsen Aug 3 '12 at 13:51
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@HaraldHanche-Olsen: Oops, yes, I did. –  Nate Eldredge Aug 3 '12 at 14:25

@Herald,

How can you be sure it has to be countable intervals? I'm not sure if that's so evident as it looks.

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