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

Let A be a measurable bounded subset of ℝ n. Given any unit vector n^∈S n-1 and s∈ℝ, there is a unique n-1 dimensional hyperplane normal to n^ and containing s⁢n^.

Define f: S n-1×ℝ→[0,∞) by sending (n^,s) to the measure of the subset of A lying on the side of the plane corresponding to (n^,s) in the direction in which n^ points. Note that (n^,s) and (-n^,-s) correspond to the same plane, but to different sides of the plane, so that f⁢(n^,s)+f⁢(-n^,-s)=m⁢(A).

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

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

Given these properties of f, we see by the intermediate value theorem that, for fixed n^, there is an interval [a, b] such that the set of s with f⁢(n^,s)=m⁢(A)/2 is [a, b]. If we define g⁢(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 ℝ. Also, since f⁢(n^,s)+f⁢(-n^,-s)=m⁢(A), if [a, b] is the interval corresponding to n^, then [-b,-a] is the interval corresponding to -n^, and so g⁢(n^)=-g⁢(-n^).

Now let A 1, A 2,…,A n be measurable bounded subsets of ℝ n, and let f i, g i be the maps constructed above for A i. Then we can define h: S n-1→R n-1 by:

| h⁢(n^)=(f 1⁢(n^,g n⁢(n^)),f 2⁢(n^,g n⁢(n^)),…⁢f n-1⁢(n^,g n⁢(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 n^∈S n-1 with h⁢(n^)=h⁢(-n^), ie, with:

| f i⁢(n^,g n⁢(n^))=f i⁢(-n^,g n⁢(-n^))=f i⁢(-n^,-g n⁢(n^)) |

where we’ve used the property of g mentioned above. But this just means that for each A i with 1≤i≤n-1, the measure of the subset of A i lying on one side of the plane corresponding to (n^,g n⁢(n^)), which is f i⁢(n^,g n⁢(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⁢(-n^,-g n⁢(n^)). In other words, the plane corresponding to (n^,g n⁢(n^)) bisects each A i with 1≤i≤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.