# regular Borel measure and regular signed measure

I was working on these two questions for a while, I think they are kind of related but I could not figure out any of them:

1)Let $\mu$ be a regular Borel measure on $[0,1]$ such that $\int_{0}^{1}{x^{2n}d\mu}=0$ for all $n=0,1,2,...$. Prove that $\mu=0$.

2) Describe all regular signed measures $\mu$ on $[-1,1]$ such that $\int_{-1}^{1}{x^{2n+1}d\mu}=0$ for all $n=0,1,2,...$.

My attempt: For the fist one, I tried using this fact that as a consequence of Stone-Weierstrass Theorem, every continuous function can be uniformly approximated by an even polynomial and then I tried to use Riesz-Markov Theorem, then I think I lost there

Recalling Riesz-Markov Theorem: Let $X$ be a locally compact Hausdorff space and $T$ a positive linear functional on $C_c(X)$, then there is a unique Radon measure (regular Borel measure) $\mu$ on $B(X)$, the Borel $\sigma$-algebra associated with the topology on $X$, for which $T(f)=\int_{X}{fd\mu}$ for all $f\in{C_c(X)}$.

Any hint or comment would be highly appreciated.

• You can extend $F_\mu:C_c([0,1]) \rightarrow \mathbb{R}$ to $F^*:C([0,1]) \rightarrow \mathbb{R}$ from Hahn-Banach theorem. From assumption $F^* \equiv 0$ on polynomial subspace (dense) of $C([0,1])$ , then this would imply $F^*\equiv 0$, and therefore $F_\mu\equiv 0$ . – Xiao Apr 26 '16 at 19:50
• @Xiao: but if $F_{\mu}$ is identically zero, does it mean $\mu=0$? – Mary Apr 26 '16 at 21:10
• Yes. Easier to think about it backward, the zero measure $\mu\equiv 0$ gives you the zero continuous linear functional from integration, by uniqueness, the zero continuous linear functional has to correspond to the zero measure. – Xiao Apr 27 '16 at 0:22
• I see what you mean. Thank you! – Mary Apr 27 '16 at 14:35
• @Mary Out of interest, where do the problems come from? – Milen Ivanov Apr 28 '16 at 19:51

Part 1 has been answered in the comments. The classification I got for part 2 is:

$\mu$ satisfies the given condition if and only if $\mu$ is "even", that is, for every Borel set $E \subset [-1;1]$ $\mu(E) = \mu(-E)$

Proof: Assume $\mu(E) = \mu(-E)$ for every Borel $E$. Let $f_n$ be simple functions which increase to $x^{2n+1}$, as defined on the interval $[0;1]$ Extend $f_n$ to odd functions on $[-1;1]$ As $f_n$ are simple and $\mu(E) = \mu(-E)$ it follows $\int f_n d\mu = 0$ so by the Dominated convergence theorem $\int _{-1}^1 x^{2n+1} d\mu = 0$, as required. (we can even write $f_n$ explicitly but there is no need)

Now, assume $\int _{-1}^1 x^{2n+1} d\mu = 0$ for all $n$ First, we will show that every odd continuous function on $[-1;1]$ can be approximated uniformly by an odd polynomial. Let $f$ be an odd function and $p_n$ be a sequence of polynomials, converging uniformly to $f$, as given by the Stone-Weierstrass theorem. Then $\frac{p_n(x) - p_n(-x)}{2} \to \frac{f(x) - f(-x)}{2} = f(x)$, since $f$ is odd. \frac{p_n(x) - p_n(-x)}{2} is also an odd polynomial, which we wanted to show.

Note that every odd polynomial is of type $a_k x^{2k+1} + a_{k-1}x^{2k-1} + ... + a_1x$, so such a polynomial integrates to 0 with respect to$\mu$ As a corollary of the uniform convergence we established, for every odd function $f$, we have $\int f d\mu = 0$

Finally, fix a Borel set $E \subset [0;1]$ We want to show $\mu (E) = \mu(-E)$ and evidently there is no loss of generality in assuming that $E \subset [0;1]$ only. Find a compact set $K$ and an open set $U$ such that $K \subset E \subset U$ and $\mu(U-K) < \epsilon$. By Urysohn's lemma there exists a continuous function $f_\epsilon : [0;1] \to [0;1]$ which has value $1$ on $K$ and $0$ on $U^c$. Thus, we have approximated $\mathbb{1}_E$ by a continuous function and $| \int_0^1 \mathbb{1}_E - f_\epsilon d\mu| < \epsilon$ Extending all functions to odd functions on $[-1;1]$ gives that $|\int_{-1}^1 \mathbb{1}_E - \mathbb{1}_{-E} - f_\epsilon d\mu| < 2\epsilon$ Letting $\epsilon \to 0$ and recalling that $f_\epsilon$ integrates to $0$ gives

$\int \mathbb{1}_E - \mathbb{1}_{-E} d\mu = 0$ or, in other words $\mu(E) = \mu(-E)$ which we wanted to prove.