Let $\Omega$ be a set, $(\Omega',\mathscr{F}')$ a measurable space, and $$f_\lambda: \Omega \to \Omega',\quad \lambda \in \Lambda$$ functions indexed by an arbitrary set $\Lambda$. Let $\mathscr{F}$ be the intersection of all $\sigma$-algebras $\mathscr{D}$ of $\Omega$ such that $f_\lambda: (\Omega, \mathscr{D}) \to (\Omega',\mathscr{F}')$ is measurable for every $\lambda \in \Lambda$.Prove that
$\mathscr{F} = \sigma(\{f_\lambda^{-1}(B):\;\lambda \in \Lambda,\;B \in \mathscr{F}'\})$.
I believe I understand what is stated here. Could you give me any tips to start proving this?
Since $f_\lambda ^{-1}(B)\in \mathscr F$ for all $\lambda $ and all $B\in \mathscr F'$, $$\sigma \{f_\lambda ^{-1}(B)\mid \lambda \in \Lambda , B\in \mathscr F'\}\subset \mathscr F.$$ The other inclusion is by definition of $\mathscr F$.