```
\begin{theorem}[General Brunn-Minkowski inequality in $\mathbb{R}^n$]
Let $0< \lambda <1$ and let X and Y be nonempty bounded measurable sets in $\mathbb{R}^n$ such that $(1- \lambda)X + \lambda Y$ is also measurable. Then,
\begin{equation} \label{eq:8}
\mu ((1- \lambda )X+ \lambda Y)^{\frac{1}{n}} \geq (1- \lambda) \mu(X)^{\frac{1}{n}} + \lambda \mu(Y)^{\frac{1}{n}}
\end{equation}
where $\mu$ is the Lebesgue measure.
\end{theorem}
\begin{remark}
We note that in general the assumption that the set $(1-\lambda)X+ \lambda Y$ is measurable is necessary even given that both $X$ and $Y$ are measurable. However, for borel sets $X$ and $Y$, we get that $(1-\lambda)X+ \lambda Y$ is a continuous image of their product and so guaranteed being measurable.
\end{remark}
```

Now we show how the
`$Brunn-Minkowski$`

theorem
directly implies the very well-know isomperimetric inequlaity:
```
\begin{corollary}[The Isoperimetric Inequality]
Consider the convex body K with surface area $S(K)$. \\
Then, the quantity $C(X)= \frac{\mu{(K)}^{1/n}}{S(K)^{1/(n-1)}}$ is maximized on Euclidean balls.
\begin{proof}
By $Brunn-Minkowski$ inequality, we have
\begin{equation}
\begin{split}
\mu{(K+ \epsilon B)} &\geq (\mu{(K)}^{1/n} + \epsilon \mu(B)^{1/n})^n \\&= \mu(K)(1+ \epsilon (\frac{\mu(B)}{\mu(K)})^{1/n})^n\\
& \geq \mu(K)(1+n \epsilon (\frac{\mu(B)}{\mu(K)})^{1/n})
\end{split}
\end{equation}
where we derived the last inequality by $(1+x)^n \geq 1+nx$ for $x \geq 0$. Now we lower bound the surface area
\begin{equation}
S(K) \geq n \mu(K)(\frac{\mu(B)}{\mu(K)})^{1/n}
\end{equation}
Now since $S(K) = n \mu(B)$ (by $Minkowski-Steiner$ formula), we get:
\begin{equation}
\begin{split}
\frac{S(K)}{S(B)}&=\frac{S(K)}{n\mu(B)}\geq \frac{\mu(K)(\frac{\mu(B)}{\mu(K)})^{1/n}}{\mu(B)}\\
&= \mu(K)^{\frac{n-1}{n}} \mu(B)^{\frac{1-n}{n}}.
\end{split}
\end{equation}
Finally, we get the isoperimetric inequality:
\begin{equation}
\frac{\mu(B)^{1/n}}{(S(K))^{1/{n-1}}} \geq \frac{\mu(K)^{1/n}}{S(K)^{1/{n-1}}}
\end{equation}
\end{proof}
\end{corollary}
\begin{corollary}[The Classical Isoperimetric Inequality in the Plane]
\begin{equation}
L^2 \geq 4 \pi A
\end{equation}
where $A$ is the area of a domain enclosed by a curve of length $L$.
\end{corollary}
```

.