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Denote by $P$ the set of all (Borel) probability measures with full support on $\left[a,b\right]$. Consider the following set: $$D \equiv \left\{f\in \mathbb{R}^{[a,b]}\vert -\infty<\int_a^b fdp<+\infty,\forall p \in P\right\}$$

My conjecture is $C[a,b] \subseteq D \subseteq B[a,b]$. Is it right?

For the part of $C[a,b] \subseteq D$, I thought every continuous function is measurable by Lusin's theorem, and then I can use dominated convergence theorem. But this thread suggests it is not the case: $f$ a real, continuous function, is it measurable? I am confused.

For the part of $D \subseteq B[a,b]$, I guess if the function is not bounded, then one can find a probability measure such that the integral explodes but I am not sure how to implement this.

Ypbor
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    To clarify why the question you linked doesn't contradict Giorgos's answer, the OP in the link was trying to show every continuous function was (Borel) measurable. In the last paragraph or so, they suggested that if they could show that every subset of $\mathbb{R}$ was measurable, they'd be set. It's true that if every subset of $\mathbb{R}$ was measurable, then every continuous function -and indeed every function period- would be measurable, but not all subsets of $\mathbb{R}$ are measurable. The answer there shows every continuous function is Borel measurable. – AJY Feb 09 '23 at 14:57
  • @AJY Thank you for the clarification. Is my following understanding correct? On the one hand, the conjecture of the OP in the link is wrong, which he/she believed to be the piece needed for the proof. On the other hand, even though the conjecture is false, the statement that continuous function is Borel measurable is true. – Ypbor Feb 10 '23 at 02:53
  • Correct. Every continuous function is Borel, as demonstrated in an answer on that post. But not every subset of $\mathbb{R}$ is Borel. – AJY Feb 13 '23 at 22:39

1 Answers1

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You can show that $D$ is exactly the set of (Borel) measurable bounded functions on $[a,b]$. Indeed, if $f$ is bounded and measurable then there exists a $C$ such that $|f(x)|\leq C$ for all $x\in [a,b]$. Then $$\left| \int_{[a,b]} f\,dp\right| \leq \int_{[a,b]}|f|\,dp \leq C$$ for all probability measures $p$, so $f\in D$

Now suppose $f$ is unbounded. Wlog we can assume that $f^+ = \max\{0, f(x)\}$ is unbounded. Then we can find a sequence $(x_n)_{n\in \mathbb{N}} \subseteq [a,b]$ with $x_n\neq x_m$ for $n \neq m$ such that $f^+(x_n) > 2^n$. Now let $$p=\sum_{n=1}^\infty \frac{1}{2^n}\delta_{x_n}$$ where $\delta_x$ denotes the Dirac measure. It's easy to see that $p$ is a probability measure on $[a,b]$ and $$\int_{[a,b]} f^+\,dp=\sum_{n=1}^{\infty}\frac{1}{2^n}\int_{[a,b]}f^+\,d\delta_{x_n} = \sum_{n=1}^\infty \frac{f^+(x_n)}{2^n} = \infty$$ so $f$ cannot have bounded expectation with respect to $p$ and thus $f\notin D$.

Now the fact that $C([a,b]) \subseteq D$ is immediate since continuous functions on $[a,b]$ are Borel measurable and bounded.

Giorgos Giapitzakis
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  • Thank you for your answer. Denote the set of bounded and measurable functions on $[a,b]$ as $BM([a,b])$. For the direction of $BM[(a,b)]\subseteq D$, am I right that one needs to first use dominated convergence theorem to ensure the integral is well-defined, and then apply your inequality? For the direction of $D \subseteq BM([a,b])$, my original post considers only those probability measure with full support on $[a,b]$, but I think the one you constructed does not satisfy this condition, right? – Ypbor Feb 10 '23 at 02:46
  • @Ypbor I don't see why you would need DCT here, since we are only dealing with probability measures. – Giorgos Giapitzakis Feb 10 '23 at 02:59
  • It may be obvious, but I just want to use DCT to make sure $\left| \int_{[a,b]} f,dp\right|$ and $\int_{[a,b]}|f|,dp$ are well defined. – Ypbor Feb 10 '23 at 03:42
  • @Ypbor You don't need DCT for that. Have a look at Theorem 2.2.11 here – Giorgos Giapitzakis Feb 10 '23 at 13:03