Identifying a multiplicative measure

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.

$$(\int f d\mu)(\int gd\mu)=0$$ 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 $\mu=\delta_{1/2}.$