$P$ 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 ?
