Let$(X,\mathcal{E},\mu)$ be a measure space, and $(B_n)\subseteq \mathcal{E}$.

I want to show that $\mu (\bigcap_{n\in \mathbb{N}} B_n )\leq \inf_{n\in \mathbb{N}} \mu(B_n)$. And I have done the following: Since $\cap_{n\in \mathbb{N}}B_n \subseteq B_n \forall n\in \mathbb{N}$ we get that $$\mu\left(\bigcap_{n\in\mathbb{N}}B_n\right)\leq \mu (B_n)$$ And now I can just insert $\inf$ on the RHS, but i dont understand why? Could anyone explain to me why this holds?

Now let $(B_n)$ be a falling sequence. Could anyone help me show that $$\mu\left(\bigcap_{n\in\mathbb{N}}B_n\right)\leq \lim_{n\rightarrow \infty}\mu(B_n)$$

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    $\begingroup$ Do you know the definition of infimum? $\endgroup$ – Shashi Dec 20 '17 at 19:41

You can insert $\inf$ in the first inequality because it holds for all $n$. For the second part, you use the first: if $(B_n)_{n \in \Bbb N}$ is decreasing, then $(\mu(B_n))_{n \in \Bbb N}$ is a decreasing sequence of real numbers, bounded below, and so $$\lim_{n\to+\infty}\mu(B_n) = \inf_{n \in \Bbb N}\mu(B_n).$$


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