Note that if c > 1 then the set SÇ[c,¥) is null. For if there is a subset of S of measure r > 0 each element of which is ³ c then for even m, your integral would be ³ r c^m which tends to ¥ as m®¥, a contradiction. This means that SÇ[1,¥) is null, by the countable union property. Similarly SÇ(-¥,-1] is null. So S is completely contained in [-1,1] apart from a null set.

Now let e > 0 and let f be a continuous function which is 1 on [-1,-e] and 0 on [0,1] (and linear in between). For any d > 0, by Weierstrass' theorem we can find a polynomial p such that |p(x)-f(x)| < d/2 for all x in [-1,1]. Therefore |q(x)-f(x)| < d for all x in [-1,1] where q(x)=p(x)-p(0). So q is a polynomial with no constant coefficient.

Now we have:

1/S ' ò_S x^m dx=ò₀¹ x^m dx

for all positive integers m. Therefore:

|1/S ' ò_S q(x) dx|=|ò₀¹ q(x) dx| < d

It quickly follows that [-1,-e]ÇS is null so [-1,0]ÇS is null. So S is completely contained in [0,1] apart from a null set.

Actually we can simplify this somewhat. As above S[-1,1] is null. Now we're given:

1/S ' ò_S x^m dx=ò₀¹ x^m dx

for all positive integers m. However this relationship is also trivially true for m=0 (a fact I managed to miss earlier!) Therefore for any polynomial P we have:

1/S ' ò_S P(x) dx=ò₀¹ P(x)dx

So by Weierstrass this is true for any continuous function P (uniformly approximate the continuous function by polynomials on [-1,1] and it doesn't matter what happens outside [-1,1]).

Certainly S is null on [-1,0] (let P(x) = 0 for positive x and P(x) = -x/e on [-e,0] and P(x) = 1 for x < -e).

So S is contained in [0,1] except for a null set and:

1/S ' ò_S P(x) dx = ò₀¹ P(x) dx

for all continuous P. Then for any subinterval [a,b] of [0,1] set P(x) = 1 on [a,b] and make it fall off to 0 very rapidly outside [a,b]. Taking the limit we have 1/S ' ò_{SÇ[a b]} 1 dx = ò_a^b 1 dx so the measure of S Ç[a,b] is S ' (b-a) as I stated earlier.

Of course we can do this last bit without using the density theorem. S has measure S ' so it has outer measure S ' . Therefore for any e > 0 we can cover S with countably many disjoint open intervals I₁, I₂, ... such that

|S ' -

å

|Ii || < e

. Therefore:

|S ' - 1/S '

å

measure S ÇIi | < e

So by the countable union property:

|S ' - 1/S ' measure (S ÇI₁ ) È(S ÇI₂) È...| < e But the set (S ÇI₁) È(S ÇI₂ ) È... is just S (all its elements are in S and all elements of S are in one of the I_is). So: