Identifying a multiplicative measure I am trying to solve the next problem:
Let $\mu:\mathcal{B}_{[0,1]}\rightarrow\mathbb{R}$ be a multiplicative measure, i.e. $$\int fg d\mu=\left(\int f d\mu\right)\left(\int g d\mu\right)$$ for all continuous $f,g:[0,1]\rightarrow\mathbb{R}.$ 
Identify $\mu.$
I was thinking in two possibly cases. One of them involves the Lebesgue measure and the other the trivial measure: If we consider the constant function $f=g=1$ we have $\mu([0,1])=\mu([0,1])^{2}$ so $\mu([0,1])=0$ or $\mu([0,1])=1.$ If  $\mu([0,1])\neq 0,$ I'm not sure $\mu$ be the Lebesgue measure, because in general, for Lebesgue measure, doesn't satisfy the equality $\mu(A\cap B)=\mu(A)\mu(B).$
Also I was considering the case $f=g$ to get $\int f^2 d\mu=\left(\int f d\mu\right)^2$ but it seems a little useless.
Any kind of help is thanked.
 A: Assume $\mu$ is finite.
Theorem. If $\int f^2 = (\int f)^2$ then $f$ is constant $\mu$-a.e.
Proof. For we have $\int (f - \int f)^2 = \int f^2 - (\int f)^2 = 0.$ Q.E.D.
Therefore, every function is constant with respect to $\mu.$ This implies $\mu$ is concentrated in one point, say $x_0.$ (To see this, simply observe the function $t \mapsto t$ is constant with respect to $\mu.$)
Observe now that if $\mu = c \varepsilon_{x_0}$ then $\int fg = c f(x_0) g(x_0)$ while $\int f \int g = c^2 f(x_0) g(x_0)$ and $c = 0$ or $c = 1.$ Clearly, the measure zero and any Dirac measure works. The end.
A: $$(\int f d\mu)(\int gd\mu)=0$$ for all $f,g\in C([0,1])$ such that $fg=0$.
So either $(\int f d\mu)=0$ or $(\int g d\mu)=0$ or both.
Take $0\leq f_1,f_2\leq1$ such that $f_1|_{[0,\frac{1}{2}]}=0$ and $f_2|_{[\frac{1}{2},1]}=0$. We have that either $\int f_1=0$ or $\int f_2 =0$ or both.
If $\int f_1\neq0$ then all the functions in $C([\frac{1}{2},1])$ have zero integral and the measure is concentrated on $[0,\frac{1}{2}]$. Similarly with $[\frac{1}{2},1]$. With a dichotomy argument you get a sequence of compact set with lenght converging to zero, thus getting a point and $\mu=\delta_x$.
If for all such $f_1$ and $f_2$ both $\int f_1=0$ and $\int f_2=0$ you then get $\mu=\delta_{1/2}.$
A: We need the following fact:

If $f$ is strictly positive on a Borel set $A$, and $\int_A fd\mu=0$, then $\mu(A)=0$.

Take some continuous positive function $f$ whose support is $[0, \frac12)$, and some continuous positive function $g$ whose support is $(\frac 12, 1]$. Their product is zero on $[0, 1]$ and so we have either $\int f=0$ or $\int g=0$. So either $\mu([0, \frac12))=0$ or $\mu((\frac12,1])=0$. Thinking along these lines, it's possible to show that $\mu$ is either zero or a Dirac measure.
