Let $X$ be a Banach space and $\mu$ be a finite measure on $X$.
If $f$ is concave on a Banach-Space $\mathbb{R}$, then $ f(tx + (1-t)y) \leq tf(x) + (1-t)f(y). $ Taking integrals on both sides implies that the functional $f\mapsto\int_{x \in X}f(x) d\mu$ is concave on $X$, by linearity of integration.
However, it is not clear to me that the same is true for quasi-concave functions $f$... That is, is $$ f\mapsto\int_{x \in X}f(x) d\mu $$ quasi-concave on $X$ is $f$ is quasi-concave?
This is not true. Take $f(x)=\sqrt{|x|}$. This is quasi-convex. Set $X=\{1,2\}$ with counting measure. Then $$ x\in \mathbb R^2 \mapsto f(x_1) +f(x_2) $$ is not quasiconvex: The points $(1,0)$, $(0,1)$ are mapped to $1$, their midpoint to $\sqrt 2$. Hence the sub-level set to level $1$ is not convex.