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.