This proof uses the Borsuk-Ulam theorem, which states that any continuous function from $S^n$ to $\mathbb{R}^n$ maps some pair of antipodal points to the same point.

Let $A$ be a measurable bounded subset of $\mathbb{R}^n$ . Given any unit vector $\hat n \in S^{n-1}$ and $s \in \mathbb{R}$ , there is a unique $n-1$ dimensional hyperplane normal to $\hat n$ and containing $s \hat n$ .

Define $f:S^{n-1} \times \mathbb{R} \rightarrow [0,\infty)$ by sending $(\hat n,s)$ to the measure of the subset of $A$ lying on the side of the plane corresponding to $(\hat n,s)$ in the direction in which $\hat n$ points. Note that $(\hat n,s)$ and $(-\hat n,-s)$ correspond to the same plane, but to different sides of the plane, so that $f(\hat n,s)+f(-\hat n,-s)=m(A)$ .

Since $A$ is bounded, there is an $r>0$ such that $A$ is contained in $\overline{B_r}$ , the closed ball of radius $r$ centered at the origin. For sufficiently small changes in $(\hat n,s)$ , the measure of the portion of $\overline{B_r}$ between the different corresponding planes can be made arbitrarily small, and this bounds the change in $f(\hat n,s)$ , so that $f$ is a continuous function.

Finally, it's easy to see that, for fixed $\hat n$ , $f(\hat n,s)$ is monotonically decreasing in $s$ , with $f(\hat n,-s)=m(A)$ and $f(\hat n,s)=0$ for $s$ sufficiently large.

Given these properties of $f$ , we see by the intermediate value theorem that, for fixed $\hat n$ , there is an interval $[a,b]$ such that the set of $s$ with $f(\hat n,s)=m(A)/2$ is $[a,b]$ . If we define $g(\hat n)$ to be the midpoint of this interval, then, since $f$ is continuous, we see $g$ is a continuous function from $S^{n-1}$ to $\mathbb{R}$ . Also, since $f(\hat n,s)+f(-\hat n,-s)=m(A)$ , if $[a,b]$ is the interval corresponding to $\hat n$ , then $[-b,-a]$ is the interval corresponding to $-\hat n$ , and so $g(\hat n)=-g(-\hat n)$ .

Now let $A_1,A_2,...,A_n$ be measurable bounded subsets of $\mathbb{R}^n$ , and let $f_i,g_i$ be the maps constructed above for $A_i$ . Then we can define $h:S^{n-1} \rightarrow R^{n-1}$ by: $$ h(\hat n) = (f_1(\hat n,g_n(\hat n)),f_2(\hat n,g_n(\hat n)),...f_{n-1}(\hat n,g_n(\hat n))) $$

This is continuous, since each coordinate function is the composition of continuous functions. Thus we can apply the Borsuk-Ulam theorem to see there is some $\hat n \in S^{n-1}$ with $h(\hat n)=h(-\hat n)$ , ie, with: $$ f_i(\hat n,g_n(\hat n))=f_i(-\hat n,g_n(-\hat n))=f_i(-\hat n,-g_n(\hat n)) $$

where we've used the property of $g$ mentioned above. But this just means that for each $A_i$ with $1 \leq i \leq n-1$ , the measure of the subset of $A_i$ lying on one side of the plane corresponding to $(\hat n,g_n(\hat n))$ , which is $f_i(\hat n,g_n(\hat n))$ , is the same as the measure of the subset of $A_i$ lying on the other side of the plane, which is $f_i(-\hat n,-g_n(\hat n))$ . In other words, the plane corresponding to $(\hat n,g_n(\hat n))$ bisects each $A_i$ with $1 \leq i \leq n-1$ . Finally, by the definition of $g_n$ , this plane also bisects $A_n$ , and so it bisects each of the $A_i$ as claimed.