Suppose $(\Omega,\mathcal{S},\mathbb{P})$ is a probability space and $\{X_i \}_{i=1,\ldots,n}$ are independent random variables with $f_k:\mathbb{R}\rightarrow\mathbb{R}$ densities and $X=(X_1,\ldots,X_n)$ the associated random vector, i'm quite sure that the following equation is true $$\mathbb{P}(X \in A ) = \int_A \prod_{k=1}^n f_k(x_k) \, d\mu^{(n)}(x_1,\ldots,x_n), \forall A \in \mathcal{B}(\mathbb{R}^n) $$
$\mu$ is the Lebesgue measure on $\mathbb{R}$
but i'm not so sure about my demonstration and i need a check:
for independence $$\mathbb{P}(X \in \prod_{k=1}^nA_k)=\mathbb{P}(\bigcap_{k=1}^nX_k \in A_k)=\prod_{k=1}^n\mathbb{P}(X_k \in A_k)= \\ = \prod_{k=1}^n \int_{A_k} f_k(x_k) \, d\mu(x_k)= \text{tonelli}= \int_{\prod_{k=1}^nA_k} \prod_{k=1}^nf_k(x_k)\,d\mu^{(n)}(x_1,\ldots,x_n)$$
the last term of che equations chain comes out from the measure $$\lambda(A)= \int_{A}\prod_{k=1}^nf_k(x_k)\,d\mu^{(n)}(x_1,\ldots,x_n)$$
moreover the product measure $\mu^{(n)}$ on $\mathbb{R}^n$ has been built considering the algebra $R$ of disjoint finite union of $A_1 \times \cdots \times A_n$, so if $A= \bigcup_{k=1}^n R_k \in R$ with $R_k =A_{1,k} \times \cdots \times A_{n, k} $ pairwise disjoint, we have $$\mathbb{P}(X \in \bigcup_{k=1}^n R_k)= \mathbb{P}(\bigcup_{k=1}^n X \in R_k)$$ the sets $X \in R_k$ are also disjoint, then
$$\mathbb{P}(\bigcup_{k=1}^n X \in R_k)=\sum_{k=1}^n \mathbb{P}(X \in R_k)= \sum_{k=1}^n \lambda(R_k)= \lambda(\bigcup_{k=1}^n R_k)$$
so we have two measures on $\mathbb{R}^n$: $\mathbb{P}_X$ and $\lambda$ which are equals on the algebra $R$
now the claim will be obtained demostrating that $\mathbb{P}_X = \lambda$ everywhere. I'll use a parallel result to conclude: suppose that $\mathcal{S} \subset \mathcal{P}(X)$ is an algebra, $\lambda$ and $\mu$ are finite measures(or $\sigma$-finite) on a measurable space $(X,\sigma(\mathcal{S}))$ and the set $M= \{ A \in \sigma(\mathcal{S}): \lambda(A) = \mu(A) \}$ contains $\mathcal{S}$ then $M = \sigma(\mathcal{S})$
so because $\lambda=P_X$ on $R$ they are equals on $\sigma(R)$ which is the $\sigma$-algebra product