For each $n \ge 1$, let $\Gamma_{n} \subset \mathbb{R}^{n}$ be a (non-empty) measurable and bounded set and equip $\Gamma_{n}$ with its Borel $\sigma$-algebra $\mathbb{B}_{n}$. Define $\Omega_{n} := \Gamma_{n} \times \{n\}$ and $\Omega := \bigcup_{n=1}^{\infty}\Omega_{n}$. For each $E \subset \Omega$, set: $$E_{n}:= \{ x \in \bigcup_{n=1}^{\infty}\Gamma_{n} : (x,n) \in E\}$$ We equip $\Omega$ with the $\sigma$-algebra: $$\mathcal{F} = \{E \subset \Omega: \mbox{$E_{n} \in \mathbb{B}_{n}$ for all $n$}\}$$
If we equip each $(\Omega_{n},\mathbb{B}_{n})$ with a measure, say, $\mu_{n}$, we can define a new measure $\mu$ on $(\Omega, \mathcal{F})$ by setting: $$\mu = \sum_{n=1}^{\infty}\mu_{n}$$
Suppose each $\mu_{n}$ has density $\rho_{n}$ with respect to the standard Lebesgue measure $\lambda_{n}$ on $(\Gamma_{n},\mathbb{B}_{n})$, that is, $d\mu_{n} = \rho_{n}d\lambda_{n}$. Of course, one can define a new function $\rho$ on $\Omega$ by setting $\rho(x,n) := \rho_{n}(x)$. We can also let $\lambda$ be the associated measure: $$\lambda = \sum_{n=1}^{\infty}\lambda_{n}$$
Question: Is it true that $\mu$ has density $\rho$ with respect to $\lambda$? In other words, does $$\int_{\Omega}f(x)d\mu(x) = \int_{\Omega}f(x)\rho(x)d\lambda(x) = \sum_{n=1}^{\infty}\int_{\Omega_{n}}f(x,n)\rho_{n}(x)d\lambda_{n}(x)$$ hold?
Solution:
We want to determine whether the measure $\mu$ has density $\rho$ with respect to the measure $\lambda$. In other words, we want to verify whether the equation $$\int_{\Omega} f(x) \, d\mu(x) = \int_{\Omega} f(x) \rho(x) \, d\lambda(x)$$ holds.
Let's break down the integral on the left-hand side: $$\int_{\Omega} f(x) \, d\mu(x) = \sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \, d\mu(x)$$
Since $\mu = \sum_{n=1}^{\infty} \mu_{n}$, we can express the above integral as: $$\sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \, d\mu(x) = \sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \, d(\mu_{n})$$
Using the fact that $\mu_{n}$ has density $\rho_{n}$ with respect to $\lambda_{n}$, we can rewrite the above as: $$\sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \rho_{n}(x) \, d\lambda_{n}(x)$$
Now, on the right-hand side of the original equation, we have: $$\int_{\Omega} f(x) \rho(x) \, d\lambda(x) = \sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \rho(x, n) \, d\lambda_{n}(x)$$
Since $\rho(x, n) = \rho_{n}(x)$, we can simplify this to: $$\sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \rho_{n}(x) \, d\lambda_{n}(x)$$
As we can see, the expressions on both sides of the original equation are the same: $$\sum_{n=1}^{\infty} \int_{\Omega_{n}} f(x, n) \rho_{n}(x) \, d\lambda_{n}(x)$$
Therefore, we can conclude that the equation $$\int_{\Omega} f(x) \, d\mu(x) = \int_{\Omega} f(x) \rho(x) \, d\lambda(x)$$ does indeed hold, and the measure $\mu$ has density $\rho$ with respect to the measure $\lambda$.
This establishes the desired result.
I put in my best effort to solve the math problem, but I acknowledge that there might be some mistakes. If anyone identifies the errors or can provide corrections, I would greatly appreciate it. Thank you for your understanding and assistance.