minimization over all probability distributions Let $$\mathcal{M} \equiv \Bigl\{\text{probability measure } \mu \text{ over } [0,1] \Big| \int_0^1 x \mathrm{d} \mu = 1/2 \Bigr\}.$$ Let $\underline{x} \in (0,1)$ be a fixed number. If $\underline{x} > 1/2$, it is easy to see $$\inf_{\mu \in \mathcal{M}} \int_{\underline{x}}^1 x \mathrm{d} \mu = 0.$$ But what is $$\inf_{\mu \in \mathcal{M}} \int_{\underline{x}}^1 x \mathrm{d} \mu$$ if $\underline{x} < 1/2$?
Thank you very much.

The $\int_a^b f(x)\; d\mu(x)$ notation is not good if $\mu$ is a measure that might give 
nonzero mass to one of the endpoints.  It's better to use $\int_A f(x)\; d\mu(x)$ for an integral over the set $A$, or $\int \chi_A(x) f(x)\; d\mu(x)$ where $\chi_A$ is the indicator function of $A$.
In this case, note that $$\chi_{[\underline{x}, 1]}(x) x \ge \frac{x - \underline{x}}{1-\underline{x}} \ \text{for}\ 0 \le x \le 1$$
so that
$$ \int \chi_{[\underline{x}, 1]}(x) x \; d\mu(x) \ge \int \frac{x - \underline{x}}{1-\underline{x}}\; d\mu(x) =  \frac{1/2 - \underline{x}}{1-\underline{x}}$$
and this is best possible, as you see with measures of the form 
$$ \frac{1}{2(1-s)} \delta_s + \frac{1-2s}{2(1-s)} \delta_1 $$
where $0 < s < \underline{x} \le 1/2$, $\delta_x$ being a unit mass at $x$.
