Suppose that for some measurable real-valued function $f$,
$$\int_A fdm \leq C,$$
where $m$ is a probability measure, for all Borel $A$, with $C$ being a constant not depending on the sets $A$. Does this imply that $f\leq C$ $m$-a.e.?
A similar results states that if $\int_A fdm = \int_A gdm,$ then $f= g$ $m$-a.e. (e.g. Is $f=g$ almost everywhere if their integrals over any subset are the same?), but I believe one cannot simply extend this result to this problem. I made an attempt by considering sets like $A_n = \{x:f(x) – 1/n \leq C\}$, so that one wants to show that $m(\cup A_n)=1$, but I wasn't able to going forward meaningfully.
Also, what if e.g. $m$ has a density, $f$ is continuous or anything? Is this result true at least in some cases? Any book that might contain this sort of results? Any help is appreciated.
Best Answer
Note that for non-negative $f,$ your hypothesis boils down to $\int fdm\le C.$
So: no, it is too weak to imply $f\le C$ a.e., even for the simplest measured space, $m(\{0\})=m(\{1\})=\frac12$: take $f(0)=3,f(1)=1,C=2.$
And you can easily transform this counterexample to have a density for $m$ and continuity for $f.$
The correct hypothesis to get $f\le C$ a.e. is $\int_Afd m\le Cm(A)$ for all measurable $A.$