# right continuous continuous function is measurable

Let $f: S \times [0, \infty)\rightarrow \mathbb{R}$ satisfy $f(x, t)$ is continuous in $x$ for each $t$ and right continuous in $t$ for each $x \in S$. Here $S$ is a metric space. Why is $f$ Borel measurable? (In the joint sense) I am vaguely familiar with the idea of proof I did a long time ago for if $S$ is instead a Euclidean space, and we have full continuity instead of only right. Then one interpolated. But you can't do that here anyway, since $S$ is not Euclidean, and nevermind the fact that you only have right continuity anyway. I also tried to write $f$ as a limit of joint measurable functions using the right continuity, but failed.

## 2 Answers

Edit: The stuff in quotes is not true. For example, there are bijections of reals whose graphs are non measurable. Thanks to Jeff for pointing this out.

"Suppose $S$ is separable locally compact metric space and $Y = [0, \infty)$ under right open topology. Let us show that if every vertical and horizontal section of $B \subseteq S \times Y$ is open then $B$ is Borel."

Fix $f: [0, \infty) \times S \rightarrow \mathbb{R}$. Define $f_n(x, t) = f(m/n, t)$ where $m, n \geq 1$ are such that $(m-1)/n \leq x < m/n$. Then $f_n$'s are all continuous. Now observe that for fixed $(x, t) \in [0, \infty) \times S$, as $n \rightarrow \infty$, $f_n(x, t) \rightarrow f(x, t)$.

• I am pretty lost on your proof as to why $A_D$ is open. Here are some of the things I'm sure I don't understand: Why is the inf of the $r(x)$ strictly positive? There's sequential compactness, but $r(x)$ is just a point function that is pointwise positive. Also, moving past that for a moment why would knowing $r(x)$ is bounded away from 0 tell us that $A_D$ is open, as opposed to just that it contains an open set? Thanks. – Jeff Jul 3 '13 at 23:34
• I tried another approach. Let me know if this is okay. – hot_queen Jul 4 '13 at 6:07
• looks right to me thanks. A feature of this that caught my attention is that no assumptions were required on $S$ and that this seems like it leads to a proof of Rudin's claim mentioned in Daniel Fischer's answer, but also in a special subcase. – Jeff Jul 4 '13 at 18:21
• @DanielFischer Are you sure? They are locally constant, and continuous because the topology is the RC one. Incidentally, I just found a hole in my own understanding that I patched up. I previously assumed that the borel sigma algebra of the RC topology was the same because I saw that I could use each generating set to obtain the other. But this manipulation is insufficient since maybe the RC topology isn't second countable. So actually what needs to be done is to see that the RC topology has open sets that are a countable union of half open positively distanced intervals. – Jeff Jul 4 '13 at 18:55
• @Jeff Ah, missed the topology. With that, they are indeed continuous. – Daniel Fischer Jul 4 '13 at 18:59

For a separable metric space $S$, we can use the

### Lemma:

Let $S$ be a separable metric space, $Y$ a topological space, and $f \colon S\times Y \to \mathbb{R}$ a function such that

1. $s \mapsto f(s,y)$ is continuous for each $y \in Y$, and
2. $y \mapsto f(s,y)$ is (Borel) measurable for each $s \in S$.

Then $f$ is measurable.

### Proof:

Let $D = \{s_\nu\colon\nu\in\mathbb{N}\}$ be a countable dense subset of $S$.

For $n \geqslant 1$, let $\mathfrak{U}_n = \{B(\frac1n,\,s)\colon s \in D\}$. Since $D$ is dense, $\mathfrak{U}_n$ is an open covering of $S$. Metric spaces are paracompact, hence there is a subordinate partition of unity

$$\{\varphi_{n,\nu} \in \mathscr{C}(S,\mathbb{R}) \colon 0 \leqslant \varphi_{n,\nu}(s) \leqslant 1,\, \operatorname{supp} \varphi_{n,\nu} \subset B(\frac1n,\, s_\nu)\},\quad \sum_{\nu \in \mathbb{N}} \varphi_{n,\nu} \equiv 1,$$

with the family $\{\operatorname{supp}\varphi_{n,\nu}\colon \nu\in\mathbb{N}\}$ locally finite.

Let $g_{n,\nu}(s,y) = \varphi_{n,\nu}(s)\cdot f(s_\nu,\, y)$ and $g_n(s,\,y) = \sum\limits_{\nu \in\mathbb{N}} g_{n,\nu}(s,\,y)$.

$g_{n,\nu}$ is measurable (as the product of the measurable functions $\varphi_{n,\nu} \circ \pi_S$ and $f(s_\nu,\,\cdot) \circ \pi_Y$), and $g_n = \lim\limits_{k\to\infty} \sum\limits_{\nu = 0}^k g_{n,\nu}$ is the pointwise limit of measurable functions, hence measurable (since the family of supports is locally finite, the sum contains only finitely many non-zero terms at each point, so convergence is assured).

Now, $f = \lim\limits_{n\to\infty} g_n$ shows that $f$ is measurable.

To see the latter limit, fix $(s,\,y) \in S\times Y$, and $\varepsilon > 0$. Since $f(\cdot,\,y)$ is continuous, there is an $n \in \mathbb{Z}^+$ with $d(s,\,t) < \frac1n \Rightarrow \lvert f(t,\,y) - f(s,\,y)\rvert < \varepsilon$. Then

$$\lvert g_n(s,\,y) - f(s,\,y)\rvert = \left\lvert\sum_{\nu \in \mathbb{N}} \varphi_{n,\nu}(s)\cdot\bigl(f(s_\nu,\,y) - f(s,\,y)\bigr) \right\rvert \leqslant \sum_{\nu \in \mathbb{N}} \varphi_{n,\nu}(s)\cdot \lvert f(s_\nu,\,y) - f(s,\,y)\bigr\rvert < \varepsilon$$

since whenever $\varphi_{n,\nu}(s) \neq 0$, we have $d(s,\,s_\nu) < \frac1n$ and hence $\lvert f(s_\nu,\,y) - f(s,\,y)\bigr\rvert < \varepsilon$.

With the above lemma, all that remains to be shown is that a right-continuous function $h \colon [0,\,\infty) \to \mathbb{R}$ is measurable.

Let $t_n(x) = (\lfloor 2^nx\rfloor + 1)/2^n$, and $h_n(x) = h(t_n(x))$. Then $t_n$ (and hence $h_n$) is constant on each interval $[m/2^n,\, (m+1)/2^n)$, and $t_n(x) \searrow x$ monotonically. Therefore $h_n$ is measurable, and the right-continuity of $h$ implies that $h_n \to h$ pointwise.

For the general case where $S$ is not assumed separable:

Lebesgue proved that every separately continuous function $f\colon \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ is a pointwise limit of continuous functions. W. Rudin extended this by showing that if $X$ is a metric space, then for any topological space $Y$, every separately continuous function $f \colon X \times Y \to \mathbb{R}$ is a pointwise limit of continuous functions.

(From the abstract of Maxim R. Burke, Borel measurability of separately continuous functions, in "Topology and its Applications", 129, 1.)

I don't have a proof of Rudin's result handy, but, accepting that, we know that a separately continuous function $f \colon X \times Y \to \mathbb{R}$ is Borel measurable.

Now, don't equip the interval $[0,\,\infty)$ with the standard topology $\mathcal{T}_s$, instead use the topology $\mathcal{T}_h$ generated by the half-open intervals $[a,\,b)$.

Take that as the space $Y$. A function $g \colon \bigl([0,\,\infty),\, \mathcal{T}_h\bigr) \to \mathbb{R}$ is continuous if and only if $g \colon \bigl([0,\,\infty),\, \mathcal{T}_s\bigr) \to \mathbb{R}$ is right-continuous.

So by that result, your

$$f \colon S \times Y \to \mathbb{R}$$

is Borel-measurable.

But the Borel $\sigma$-algebra generated by $\mathcal{T}_h$ is the same as that generated by $\mathcal{T}_s$.

• Thank you for your response. However, I am looking for a proof that proceeds from elementary analysis. This is a result that is being assumed in Liggett's textbook, and it should be able to be handled without assuming any black boxes, or so I hope. In his situation $S$ is actually a separable locally compact metric space. – Jeff Jul 3 '13 at 19:39
• Ah, "a separable locally compact metric space" makes it relatively easy, I think. Not so easy that I can write down a proof out-of-hand, but I think I can find one in reasonable time. – Daniel Fischer Jul 3 '13 at 19:42
• @Jeff Separability is what yields a simple enough proof. I haven't seen a way to use local compactness, that led me down a couple of dead ends in my search. – Daniel Fischer Jul 4 '13 at 8:57
• I will study your solution in detail when I get a chance. For now, recalling from a vague memory, I think that "metric spaces are paracompact" is a very high-machinery theorem. On the other hand, a partition of unity by continuous compactly supported functions can be explicitly constructed when you have locally compact separable. – Jeff Jul 4 '13 at 15:24
• Actually, this can be problematic if I understand correctly because the covering you give is not actually locally finite. There is a locally finite partition of unity subordinate to it, but then the index set is not $\mathbb{N}^2$ which is a required assumption in how you manipulate the sums and limits. – Jeff Jul 4 '13 at 17:56