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The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.

The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we will have a set in $\mathcal{U}$ with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.

The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the outer density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive measure, then we can split $U$ in half $U=A\sqcup B$ each with half the outer density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the measure of $U$. By iterating this, we can make the outer measure of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.

The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.

The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the outer density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive measure, then we can split $U$ in half $U=A\sqcup B$ each with half the outer density. One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the measure of $U$. By iterating this, we can make the outer measure of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.

The answer is: zero.

The reason is that every ultrafilter has zero as the infimum of the outer density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive measure, then we can split $U$ in half $U=A\sqcup B$ each with half the outer density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we $\mathcal{U}$ will have a set with half the measure of $U$. By iterating this, we can make the outer measure of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.