The notion of density of distribution
This notion seems to be a measure of uniformity. If $\sigma=1$ you get a uniform distribution on $\{0,1\}^n$ (since the sum over all strings of length $n$ is $1$), but for other $\sigma$s (e.g. $\sigma = 2^{-k}$) it's still  distribution on $\{0,1\}^n$, not on $\{0,1\}^{n-k}$ or similar. Thus saying that a distribution on another (disjoint) set is an upper bound is absolutely wrong (but they are related, s.b.).

If for some $x$ the probability is greater than $\frac{1}{2^n}$ it has to be smaller for others. On the other hand for $\sigma=\frac{1}{2}$ at least half of the strings have nonzero probability since the sum of any number of elements $&lt; 2^{n-1}$ is smaller than $1$. This holds for any $\sigma &gt; 0$, i.e. $\sigma 2^n$ elements have nonzero probability.

Using this you can compare the distribution for $\sigma=2^{-k}$ to the distribution for $\sigma=1$ (i.e. uniform) on $\{0,1\}^{n-k}$. It's easy to see that the entropy of the distribution with $\sigma=2^{-k}$ on $\{0,1\}^n$ (let $H$ be a random variable with this distribution) is at least the entropy of the uniform distribution on $\{0,1\}^{n-k}$ (RV denoted by $U$) since for every injection $f: \{0,1\}^{n-k}\rightarrow \{0,1\}^n:$ 
$$\mathbb{P}(U=x)\geq \mathbb{P}(H=f(x)) \quad .$$