# Have I proved that $\sigma(\sum^n_{i=1}X_i)\subset \sigma(\bigcup^n_{i=1}\sigma(X_i))$?

Assuming the $X_i$ 's are measurable functions, how does one prove that $\sigma(\sum^n_{i=1}X_i)\subset \sigma(\bigcup^n_{i=1}\sigma(X_i))$?

I'm thinking of $\sum^n_{i=1}X_i= g(X_1,...,X_n)$, where $g$ is continuous, hence measurable.

So, for $(X_1,...,X_n)^{-1}(\prod_{i=1}^n I_i)=\bigcap X_i^{-1}(I_i)$ and $\sigma(\{I: I=(a,b), a,b \in \mathbb{R}\})=\mathcal{B}$ we have that $$\sigma((X_1,...,X_n))=\sigma((X_1,...,X_n)^{-1}(\{\prod_{i=1}^n I_i: I_i=(a_i,b_i), a_i,b_i \in \mathbb{R}\}))= \sigma(\bigcap_{i=1}^n X_i^{-1}(\{I_i: I_i=(a_i,b_i), a_i,b_i \in \mathbb{R}\})) \subset \sigma(\bigcup_{i=1}^n X_i^{-1}(\mathcal{B}))$$

Any help would be appreciated.

• How do you define the sum of your sets $\sum^n_{i=1}X_i$? Oct 7, 2017 at 12:38
• @Julien I'm not sure I understand what you're asking. The X_i are measurable functions. Oct 7, 2017 at 12:46
• @Julien I've edited the question. I've found a some typos. sorry about that. Oct 7, 2017 at 12:50
• What does $\sigma$ denote in the title? On the LHS you apply it to a sum of functions, on the RHS you apply it alternatively to a function and then to a union of sets? Oct 7, 2017 at 13:04
• @Julien $\sigma(\sum X_i)=(\sum X_i)^{-1}(\mathcal{B})$. The sigma is for sigma-algebras. Oct 7, 2017 at 13:10

I believe a proof can be derived by induction:

1/ Assume the result holds until $n=N$. Then considering $N+1$ measurables functions $X_{i}$ we have:

$\sigma(\sum_{i=1}^{N+1}X_{i}) = \sigma(\sum_{i=1}^{N}X_{i}+X_{N+1}) \subset \sigma(\sigma(\sum_{i=1}^{N}X_{i}) \cup \sigma(X_{N+1})) \subset \sigma(\sigma(\cup_{i=1}^{N}\sigma(X_{i}))\cup \sigma(X_{N+1}))=\sigma(\cup_{i=1}^{N+1}\sigma(X_{i}))$

i.e. the result holds until N+1.

2/ Now all you have to do is to prove the result for N=2. In order to do this you can show that:

$(X_{1}+X_{2})^{-1}(]-\infty;a[) = \{X_{1} < a - X_{2}\} = \cup_{q \in \mathbb{Q}}\{X_{1}^{-1}(]-\infty;q[) \cap (a-X_{2})^{-1}(]q; +\infty[)\} \subset \sigma(X_{1}) \cup \sigma(X_{2}) = \sigma(\cup_{i=1}^{2}X_{i})$ (we are using the fact that $\sigma(\{]-\infty;a[; a \in \mathbb{R}\})= \mathcal{B})$

• Julien, thanks for the answer. But why did you say that my proof was incorrect? where was the mistake? Oct 7, 2017 at 14:35
• Your mistake is that you are looking into $\sigma(X_1,...,X_n)$ while you should be looking into $\sigma(g(X_1,...,X_n))$, which means that you should be inquiring into $(g(X_1,...,X_n))^{-1}(...)$ rather than $(X_1,...,X_n)^{-1}(...)$. Your sum (i.e. function $g$) somehow "disappeared" Oct 7, 2017 at 14:44
• But we know that $(g(X_1,...,X_n))^{-1}(...) \subset (X_1,...,X_n)^{-1}(...)$ Oct 7, 2017 at 16:45
• Sorry, I hadn't seen the transition... and had indirectly reproved the fact that the summation is continuous. Your proof is correct! Oct 7, 2017 at 19:36