A Borel measure is a measure on the Borel $\sigma$-algebra that assigns a finite value to every compact subset. Let $\mu$ be a Borel measure on $[0,1]$ (with the standard topology) and assume that $$\int fg \, d\mu = \int f \, d\mu \cdot \int g \, d\mu$$ for all continuous $f,g : [0,1] \to \mathbb{C}.$ Prove that $\mu = \delta_a$ for some $a \in [0,1]$, where $\delta_a$ is the Dirac measure centered at $a$, i.e., $\delta_a(A) = \chi_A(a).$
I have not really found a good way to attack this problem. The furthest I have got is that we might be able to show that $$\mu(\{x\}) = \mu ([0,x]) \cdot \mu ([x,1])$$ by applying the given to $\chi_{[0,a]}$ and $\chi_{[a,1]}.$ Of course, these are not continuous, but I'm trying to first solve without the continuous assumption, thinking that we can probably approximate the functions in $\mathcal{L}^1(\mu)$ arbitrarily well by continuous functions.
Thanks in advance!
Taking $f=g=1$ we see that $\mu [0,1]=0$ or $1$. Assuming that $\mu$ is not the zero measure we see that $\mu$ is a probability measure. Taking $f(x)=g(x)=x$ we get $(\int xd\mu(x))^{2}=\int x^{2}d\mu(x)$. By the condition for equality in Holder's/Cauchy-Schwarz inequality we see that $x=a$ a.e $[\mu]$ for some constant $a$ which means $\mu =\delta_a$. Ref: Rudin's RCA.