How to prove $\mathbb P(X\gt \mathbb E[X])=0\implies\mathbb P(X=\mathbb E[X])=1?$ This article contains the following (in the section "The Interface Theory of Perception", on page 4 of the .pdf):

[...] we can think of a fitness function to be a function $f :W → [0,1],$ which
assigns to each state $w$ of $W$ a fitness value $f(w).$ [...] If the true probabilities of states in the world are given by a probability measure $m$ on $W$, then one can define a new probability measure $mf$ on $W$, where for any event $A$ of $W$, $mf(A)$ is simply the integral of $f$ over $A$ with respect to $m$; $\ mf$ must of course be normalized so that $mf(W) = 1.$

Thus, $mf(W):= \int_W f\,dm=\mathbb E[f]$ (i.e. the expectation of $f$, assuming measurability), so the normalization requirement is just $\mathbb E[f]=1.$ Hence we have $f :W → [0,\mathbb E[f]],$ and therefore $m\{w:f(w)>\mathbb E[f]\}=0.$
Question: Isn't it the case that $$m\{w:f(w)>\mathbb E[f]\}=0\implies m\{w:f(w)=\mathbb E[f]\}=1?$$ In other words, haven't the authors unwittingly assumed a degenerate distribution?
These are probably elementary results that I've forgotten how to prove, but I seem to recall the following intuitive relations for any real r.v. $X$ with finite expectation: $$\mathbb P(X\gt \mathbb E[X])=0\iff \mathbb P(X\lt \mathbb E[X])=0\iff \mathbb P(X=\mathbb E[X])=1.$$
 A: $\mathbb P(X>\mathbb EX)=0$ does imply $\mathbb P(X=\mathbb EX)=1$, because $$\begin{align}\mathbb EX &= \int X = \int _{X\le \mathbb EX} X =\int_{\mathbb EX-\epsilon\le X\le \mathbb EX} X +\int_{X<\mathbb EX-\epsilon} X\\ &\le(\mathbb EX) \cdot \mathbb P(\mathbb EX-\epsilon\le X \le \mathbb EX) + (\mathbb EX-\epsilon)\mathbb P (X<\mathbb E X -\epsilon) \\ &=\mathbb EX -\epsilon\mathbb P(X<\mathbb E X-\epsilon)\end{align}$$
Hence $\epsilon\mathbb P(X<\mathbb E X-\epsilon)\le 0$ and $\mathbb P(X<\mathbb E X-\epsilon)=0$ for any $\epsilon>0$. And by continuity from below, $\mathbb P(X < \mathbb E X)=0$.
However, this is not what author means in the context. Let $(W, \mathcal B, \mathbb P)$ be a probability space, and $f:W\rightarrow \mathbb R^+$ be a non-negative function such that $0<\mathbb Ef=\int_W f<\infty$, then we may construct a new probability space $(W,\mathbb B, \mu)$, with $\mu(B)=\frac{\int_B f}{\int_W f}$. That is $\frac{f}{\int_W f}$ is the probability density of the new probability measure. In other words, probabilities are redistributed proportionally over $W$ according to $f$.
The author didn't claim that $\int_W f=1$ but one has to divide it to get a probability.
