I just read a paper by Abner Shimony (The Status of the Principle of Maximum Entropy), and he is making a claim about a Lebesgue integral in appendix A that I don't fully understand yet.
Let $f(x)$ be a continuous function which has strictly one extremum at $x=0$ (it's a maximum, but according to Shimony that doesn't matter). We also know that
$$ f(0)=1/4 $$
and
$$ \int_{\mathbb{R}}f(x)d\mu=1/4 $$
where $\mu$ is a probability measure. Shimony says that it follows that $\mu$ puts all of its weight on the point $0$, i.e. $\mu(\{0\})=1$. Why, and how can you show this to be true?
$f$ has a strict maximum at $x = 0$, so $f(x) < 1/4$ for $x \ne 0$.
Let $L_n = \{x: f(x) < 1/4 - 1/n \}$. Then $$\frac{1}{4} = \int_{\mathbb R} f(x)\; dx = \int_{L_n} f(x)\; d\mu + \int_{\mathbb R \backslash L_n} f(x)\; d\mu \le \left(\dfrac{1}{4} - \dfrac{1}{n}\right) \mu(L_n) + \dfrac{1}{4} (1 - \mu( L_n)) = \frac{1}{4} - \frac{\mu(L_n)}{n}$$ so $\mu(L_n) = 0$. Now by countable additivity $$\mu(\mathbb R \backslash \{0\}) = \mu\left( \bigcup_{n \in \mathbb N} L_n \right) = 0$$