Are these functions bewteen continuous and bounded functions? 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.
 A: 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.
