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

Now let $\epsilon >0$ and let $f$ be a continuous function which is 1 on $[-1,-\epsilon ]$ and 0 on $[0,1]$ (and linear in between). For any $\delta >0$, by Weierstrass' theorem we can find a polynomial $p$ such that $|p(x)-f(x)|<\delta /2$ for all $x$ in $[-1,1]$. Therefore $|q(x)-f(x)|<\delta$ 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:

$1/S\text{'}{\int }_S{x}^m\mathrm{dx}={\int }_0^1{x}^m\mathrm{dx}$

for all positive integers $m$. Therefore:

$|1/S\text{'}{\int }_{S}q(x)\mathrm{dx}|=|{\int }_0^{1}q(x)\mathrm{dx}|<\delta$

It quickly follows that $[-1,-\epsilon ]\cap S$ is null so $[-1,0]\cap S$ 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:

$1/S\text{'}{\int }_S{x}^m\mathrm{dx}={\int }_0^1{x}^m\mathrm{dx}$

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:

$1/S\text{'}{\int }_{S}P(x)\mathrm{dx}={\int }_0^{1}P(x)\mathrm{dx}$

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 $P(x)=-x/\epsilon$ on $[-\epsilon ,0]$ and $P(x)=1$ for $x<-\epsilon$).

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

$1/S\text{'}{\int }_{S}P(x)\mathrm{dx}={\int }_0^{1}P(x)\mathrm{dx}$

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 $1/S\text{'}{\int }_{S\cap [ab]}1\mathrm{dx}={\int }_a^{b}1\mathrm{dx}$ so the measure of $S\cap [a,b]$ is $S\text{'}(b-a)$ as I stated earlier.

Of course we can do this last bit without using the density theorem. $S$ has measure $S\text{'}$ so it has outer measure $S\text{'}$. Therefore for any $\epsilon >0$ we can cover $S$ with countably many disjoint open intervals $I_1$, $I_2$, ... such that $|S\text{'}-\sum |I_i\left|\right|<\epsilon$. Therefore:

$|S\text{'}-1/S\text{'}\sum \mathrm{measure}S\cap I_i|<\epsilon$

So by the countable union property:

$|S\text{'}-1/S\text{'}\mathrm{measure}(S\cap I_1)\cup (S\cap I_2)\cup ...|<\epsilon$ But the set $(S\cap I_1)\cup (S\cap I_2)\cup ...$ is just $S$ (all its elements are in $S$ and all elements of $S$ are in one of the $I_i$s). So: