why $\int_{s_1}\cdots\int_{s_n} f = \int_{s_1}\cdots\int_{s_n} f_1 \cdots f_n$ yields $f = f_1 \dots f_n$

Question

When I am reading Kai Lai Chung's Elementary Probability Theory. In Page 145, Chapter 5.5:

If $S_1, S_2, \cdots,S_n$ are reasonable Borel sets. and $$\int_{S_1}\cdots\int_{S_n}f_1(u_1)\cdots f_n(u_n)\text{d}u_1\cdots \text{d}u_n=\int_{S_1}\cdots\int_{S_n}f(u_1 \cdots u_n)\text{d}u_1\cdots \text{d}u_n$$

where $f$ is joint density function. and $f_1 ,\cdots,f_n$ are marginal densities.

then he said the above expression yields: $$f(u_1 \cdots u_n) = f_1(u_1)\cdots f_n(u_n)$$

I don't know why this is true. If $f \not= f_1\cdots f_n$ on some points. the above integral is also equal on any reasonable Borel Sets.

Please help me . thanks very much.

Answer 1

For the measure $\mu$ defined by \begin{align*} \mu(S_{1}\times\cdots\times S_{n})&=\int_{S_{1}\times\cdots\times S_{n}}f(u_{1},...,u_{n})du_{1}\cdots du_{n}\\ &=\int_{S_{1}\times\cdots\times S_{n}}f(u_{1})\cdots f(u_{n})du_{1}\cdots du_{n}, \end{align*} then by Radon-Nikodym Theorem, we have \begin{align*} \dfrac{d\mu}{d u}=f(u_{1},...,u_{n})=f(u_{1})\cdots f(u_{n}),~~~~\text{a.s.} \end{align*} where $u=(u_{1},...,u_{n})$.

Answer 2

If two probability measures on a product space coincide on measurable rectangles they coincide everywhere. Let $\mu$ be the measure with density $f(x_1,x_2,...,x_n)$ and $\nu$ be the one with density $f_1(x_1)f_2(x_2)...f_n(x_n)$. Then the hypothesis says these two probability measures coincide on measurable rectangle. Hence $\mu =\nu$ and their densities coincide too.