# Closedness of the set of probability measures anihilating a measurable function

Let $X$ be a compact metrizable topological space and let $f$ be a bounded, real valued, Borel function on $X$. Denoting by $P(X)$ the collection of all probability measures on $X$, consider the subset $$Z(f):= \left\{μ∈P(X): \int_X fdμ=0\right\}.$$

Question: What is the most general condition one should require of $f$ to ensure that $Z(f)$ is closed in the weak* topology of $P(X)$ (seen within the dual Banach space of $C(X)$).

For obvious reasons $Z(f)$ is closed when $f$ is continuous, but it is also closed if, say, $f$ is the pointwise limit of an increasing sequence $\{f_n\}_n$ of non-negative continuous functions (hence lower semi-continuous). This is because $$Z(f) = \bigcap_nZ(f_n),$$ so $Z(f)$ is closed.

On the other hand, if the Borel subset $A⊆X$ is not open, there exists a sequence $\{x_n\}_n$ in the complement of $A$, converging to a point $p$ in $A$. In this case the Dirac measures $δ_{x_n}$ all lie in $Z(1_A)$ (by $1_A$ I mean the characteristic function of $A$), and they converge weak*ly to $δ_p$, which isn't in $Z(1_A)$. Therefore $Z(1_A)$ is not closed!

• Another trivial example is when $f$ is any strictly positive (resp., strictly negative) Borel function, for then we have $Z(f) = \emptyset$. This shows that the answer can't really have to do with the regularity of $f$. – Nate Eldredge Jul 13 '18 at 7:05
• @NateEldredge, you are right. The answer will likely be a combination of positivity and regularity. Incidentally, the case I am mostly interested in involves a function $f$ of the form $g-h$, where both $g$ and $h$ are non-negative, $g$ is continuous, and $h$ is upper-continuous. – Ruy Jul 13 '18 at 11:50
• if one assumes $f$ to be non-negative, then I'd say that $Z(f)$ is closed if and only if the lower semicontinuous envelope of $f$ (i.e. the maximal lsc function $\leq f$) has the same zeros of $f$. Not sure if this helps for the case $g-h$ mentioned above – Nicola Gigli Jul 13 '18 at 13:44

• This is true, but I don't see how it helps answer the question? $P(X)$ isn't a vector space. – Nate Eldredge Jul 13 '18 at 7:02