Let $A$ be a single point and $B$ a unit cube in $\mathbb{R}^n$, what is then the volume $\lambda \mapsto \mathrm{Vol}\big((1-\lambda)A + \lambda B\big)$?
I am not exactly sure, what the set $(1-\lambda)A + \lambda B$ is. It seems that if I Minkowski-add a single point to any (nonempty) set, I don't change its volume. So I expect $\mathrm{Vol}\big((1-\lambda)A + \lambda B\big)=\mathrm{Vol}(\lambda B)$ - is that true?
Then $(1-\lambda)A + \lambda B \overset{?}{=}\lambda B$ seems to be just an $n$-dimensional hypercube with a side of length $\lambda$ and so the volume should be $\mathrm{Vol}(\lambda B)=\lambda^n$.
It seems to easy to be true. Is the argumentation correct?
Since volume is just Lebesgue measure, it is translation invariant, so adding a single point will not change the volume of your set. So it is true, that $Vol((1-\lambda)A + \lambda B) = \lambda^n$.
The formula with the question mark does not hold, since adding a single point translates your unit cube by that vector and therefore the sets are not equal, since they are at different positions.