Sum of independent random variables being a.s. constant implies that each of them is a.s. constant I am trying to solve the following problem:
Let $(\Omega, \mathbb{A}, \mathbb{P})$ be a probability space and $X_1, X_2, \ldots, X_n$ independent real random variables.
Prove that the sum $X_1 + X_2 + ... + \ldots X_n$ is $\mathbb{P}$-almost surely constant iff each $X_i$ is $\mathbb{P}$-almost surely constant.
Do you have any ideas or hints how to tackle with this problem?
Thanks.

It suffices to show: $X+Y=c$ almost surely $\Rightarrow X$ constant almost surely where $X,Y$ are independent random variables.
Let $\xi,\eta \in \mathbb{R}$. By the independence we have
$$\mathbb{E}e^{\imath \, (X,X+Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, \xi \cdot X} \cdot \mathbb{E}e^{\imath \, (X+Y) \cdot \eta} = \mathbb{E}e^{\imath \, \xi \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot X} \cdot  \mathbb{E}e^{\imath \, \eta \cdot Y}$$
On the other hand
$$\mathbb{E}e^{\imath \, (X,X+Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, (\xi+\eta) \cdot X} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y}$$
Since $$\mathbb{E}e^{\imath \, \eta \cdot X} \cdot  \mathbb{E}e^{\imath \, \eta \cdot Y} = \mathbb{E}e^{\imath \, (X+Y) \cdot \eta} = e^{\imath \, \eta \cdot c} (\not=0)$$ we have $\mathbb{E}e^{\imath \, \eta \cdot Y} \not= 0$ for all $\eta$. Hence we obtain from the first two equations
$$\mathbb{E}e^{\imath \, (X,X) \cdot (\eta,\xi)} = \mathbb{E}e^{\imath \, (\xi+\eta) \cdot X} = \mathbb{E}e^{\imath \, X \cdot \eta} \cdot \mathbb{E}e^{\imath \, X \cdot \xi}$$
This means that $X$ is independent of $X$ and therefore almost surely constant.
Remark In the last step we used the following theorem: Two random variables $U,V$ are independent $$\Leftrightarrow \forall \xi,\eta: \mathbb{E}e^{\imath (U,V) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, U \cdot \xi} \cdot \mathbb{E}e^{\imath \, V \cdot \eta}$$

Another approach: Let $S_n := \sum_{j=1}^n X_j$. We have $0=\mathbb{V}S_n= \sum_{j=1}^n \mathbb{V}X_j$. This implies $\mathbb{V}X_j=0$ and therefore $X_j = \mathbb{E}X_j$ a.s.. The problem is that one has to show $X_j \in L^2$ (to do these calculations).
