From measure theory volume 1 by Fremlin, exercise 111Xf:

Let $X$ be a set, $\mathcal{A}$ a family of subsets of $X$, and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that $Y\subset X$. Show that ${\{E \cap Y : E \in \Sigma}\}$ is the σ-algebra of subsets of $Y$ generated by ${\{A \cap Y : A \in \mathcal{A}}\}$.

I'm able to show that it's a $\sigma$-algebra, and clearly it contains the $\sigma$-algebra generated by ${\{A \cap Y : A \in \mathcal{A}}\}$, but I don't know how to show ${\{E \cap Y : E \in \Sigma}\}$ $\subseteq\sigma (\{{A \cap Y : A \in \mathcal{A}}\})$.

  • 2
    $\begingroup$ You are trying to show $\{E\cap Y \; : \ E \in \Sigma\} \subseteq \sigma(\{A \cap Y \; : \; A \in \mathcal{A}\})$, not what you have written in your prompt. $\endgroup$
    – Mnifldz
    Jan 28, 2015 at 18:42

2 Answers 2


Let $\mathcal A' = \left\{E\cap Y : E\in\mathcal A\right\}$ and $\Sigma' = \left\{E\cap Y : E\in\Sigma\right\}$. Let

$$\mathcal F = \left\{E\in\Sigma : E\cap Y\in\sigma(\mathcal A')\right\}$$

Then $\mathcal A\subseteq\mathcal F\subseteq\Sigma=\sigma(\mathcal A)$ and $\mathcal F$ is an $\sigma$-algebra (show this). Thus $\mathcal F =\Sigma$.

Now, for any $E\in\Sigma'$, we have $E = F\cap Y$ for some $F\in\Sigma$. But by previous result, $F\cap Y\in\sigma(\mathcal A')$ for all $F\in\Sigma$, so $E\in\sigma(\mathcal A')$. This proves $\Sigma'\subseteq\sigma(\mathcal A')$.


First note that what you want to show is not true in general. You forgot to take the $\sigma$ algebra generated by $\{A\cap Y\mid \dots\}$ instead of just $\{A\cap Y\mid \dots\}$.

To prove the corrected statement, use the good set principle, i.e. set

$$ G = \{E\in \Sigma \mid E\cap Y\in \sigma(\{A\cap Y\mid A\in \mathcal{A}\})\}. $$

Show that $G$ is a $\sigma$ algebra and that $\mathcal{A}\subset G$.

Then think about why that implies your claim.


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