"Smallest" event such that probability greater than a given value Very briefly, consider the probability space $(\mathbb R^n, \mathcal{B}(\mathbb R^n),P)$. During a problem I am studying, I came to a point where i need to compute 
\begin{equation*}
\begin{aligned}
& {\text{minimize}}
& & \text{vol}(A) \\
& \text{subject to}
& & P(A) \ge 1 - \alpha
\end{aligned}
\end{equation*}
where $\text{vol(A)}$ refers the the volume in $\mathbb R^n$ (i.e. Lebesgue measure).
My question would be, whether this is some standard question, or, if it does even make sense to consider this (unfortunately I am by far no expert in probability theory).
Any feedback is greatly appreciated.Thank you.
 A: $P$ and $\lambda$ the Lebesgue measure are both $\sigma$-finite measure, so you can use the decomposition theorem and the Radon-Nikodym theorem.
By the decomposition theorem, there exist two disjoint measurable sets $E_1$ and $E_2$, and two unique measures $\lambda_1$ and $\lambda_2$ such that:
$$\lambda = \lambda_1 + \lambda_2$$
with $\lambda_1$ absolutely continuous w.r.t. $P$ (and both $\lambda_1$ and $P$ concentrated on $E_1$) and $\lambda_2$ concentrated on $E_2$.
Using the Radon-Nikodym theorem, we then get that $\lambda_1$ has a density $\frac{d\lambda_1}{dP}$ w.r.t to $P$.
Let $A_\beta = \left\lbrace x \in E_1 \ : \ \frac{d\lambda_1}{dP}\left(x\right) \leq \beta  \right\rbrace$. $A_\beta$ is measurable and the function $f$ defined by $ \beta \mapsto P(A_\beta)$ is increasing and it goes to $1$ as $\beta$ goes to $+\infty$.
If $f$ is continuous, let $f^{-1}\left(1 - \alpha \right) = \sup_{f(\beta) < 1 - \alpha}{\lbrace \beta \rbrace } = \inf_{ f(\beta) \geq 1 - \alpha}{ \lbrace \beta \rbrace } $, your minimization problem is minized at $A_{f^{-1}\left(1 - \alpha \right)}$ which has probability $P(A_{f^{-1}\left(1 - \alpha \right)}) = 1 - \alpha$ and has volume:
$$\lambda(A_{f^{-1}\left(1 - \alpha \right)}) = \lambda \left( \left\lbrace x \in E_1 \ : \ \frac{d\lambda_1}{dP}\left(x\right) < f^{-1}\left(1 - \alpha \right)  \right\rbrace \right) $$
If $f$ is not, there is still a generalized inverse $\tilde{f}^{-1}\left(1 - \alpha \right) = \sup_{f(\beta) < 1 - \alpha}{\lbrace \beta \rbrace } = \inf_{ f(\beta) \geq 1 - \alpha}{ \lbrace \beta \rbrace } $ and the minimization problem has volume $v$ bounded by:
$$\lambda \left( \left\lbrace x \in E_1 \ : \ \frac{d\lambda_1}{dP}\left(x\right) < \tilde{f}^{-1}\left(1 - \alpha \right)  \right\rbrace \right) \leq v \leq  \lambda \left( \left\lbrace x \in E_1 \ : \ \frac{d\lambda_1}{dP}\left(x\right) \leq \tilde{f}^{-1}\left(1 - \alpha \right)  \right\rbrace \right) $$
Ok, I'm stuck there. Can one show that $f$ is always continuous ? Or if not, is it always possible to decompose the set $\left\lbrace x \in E_1 \ : \ \frac{d\lambda_1}{dP}\left(x\right) = \tilde{f}^{-1}\left(1 - \alpha \right)  \right\rbrace$ into smaller measurable sets to improve things ?
