Not true. Let $S=\{1,2,\dotsc,n\}\cup\{2n\}$. The total mass of $\bar{f}$ is $\approx 3n\cdot (1/n)=3$. Let $\mu$ be the uniform measure on $\{1,2,\dotsc,n\}$. Then the total mass of $f$ is $\approx 2n\cdot (1/n)=2$. (If you do not like $\mu(2n)=0$, make $\mu(2n)=\varepsilon$ to be very small.)

The opposite inequality also fails, as is witnessed by making $\mu(2n)$ in this example big, not small.

Not true. Let $S=\{1,2,\dotsc,n\}\cup\{2n\}$. The total mass of $\bar{f}$ is $\approx 3n\cdot (1/n)=3$. Let $\mu$ be the uniform measure on $\{1,2,\dotsc,n\}$. For this choice of $\mu$, the total mass of $f$ is $\approx 2n\cdot (1/n)+n\cdot(1/2n)=2.5$. Here $2n$ counts the elements $\{2,3,\dotsc,2n\}$ and $n$ counts the elements from $2n+1$ to $3n$. (If you do not like that $\mu(2n)=0$, then make $\mu(2n)=\varepsilon$ very small.)

The opposite inequality also fails, as is witnessed by making $\mu(2n)$ in this example big, not small.