# Prove $\mu(\varliminf _{n \rightarrow \infty} A_{n}) \leq \varliminf_{n \rightarrow \infty} \mu(A_{n})$

If $$(\Omega, \mathcal{A}, \mu)$$ is a measure space, and $$A_{n} \in \mathcal{A},$$ then $$\mu(\varliminf _{n \rightarrow \infty} A_{n}) \leq \varliminf_{n \rightarrow \infty} \mu(A_{n})$$

I tried to prove the statement by going back to the definition of liminf of a set: $$\mu(\varliminf A_n)=\mu(\bigcup^\infty_{n=1}\bigcap^\infty_{n=k}A_n)$$

As I know $$\bigcap^\infty_{n=k}A_n$$ is increasing when $$k$$ increases, I get $$\mu(\bigcup^\infty_{n=1}\bigcap^\infty_{n=k}A_n)=\lim\mu(\bigcap^\infty_{n=k}A_n)$$ but I am stucked from here. Intuitively, $$\bigcap^N_{n=k}A_n$$ is decreasing when $$N\rightarrow\infty$$ but as $$n$$ starts from $$k$$, therefore it seems $$\mu(\bigcap^N_{n=k}A_n)$$ is bounded above but how do I connect it to the $$\inf \mu(A_n)$$? On the other hand, it seems $$\bigcap^\infty_{n=k}A_n$$ is a decreasing sequence that is bounded below by $$n\geq k$$. I am so confused with the bounded above and bounded below and the infimum in this case. I hope my question make sense. Could someone please explain to me how should I look at this question and reason the statement rigorously?

• You're almost there. Note that $\bigcap_{n=k}^\infty A_n\subseteq A_k$. – Jakobian Sep 23 at 22:25
• You can see it as Fatou's lemma applied to indicator functions $1_{A_n}$. – Gae. S. Sep 23 at 22:35
• @Jakobian Yes, I noticed that I think I am having troubles with the notation. I wrote $\mu(\cap_{n\geq k} A_n)\leq\mu(A_k)=\mu(\inf_{n\geq k}A_n)=\inf_{n\geq k}\mu(A_n)$ but I don't know if it makes sense or not... – JoZ Sep 23 at 22:51
• @JoZ it doesn't really make sense. But you can take this inequality and put $\liminf$ on both sides. What do you get? – Jakobian Sep 23 at 22:55

As you have written, $$\mu(\liminf_{n\to\infty} A_n) = \lim_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right)$$ where the limit on the right exists (i.e. it is in particular equal to $$\liminf_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right)$$). As @Jakobian was saying, $$a_k:=\mu\left(\bigcap_{n=k}^\infty A_n\right) \leq \mu(A_k) =: b_k$$. Now if we have two sequences of real numbers $$\{a_k\}_{k=1}^\infty$$ and $$\{b_k\}_{k=1}^\infty$$ where each $$a_k \leq b_k$$, then the $$\liminf$$ of the $$a_k$$'s MUST be $$\leq$$ the $$\liminf$$ of the $$b_k$$'s, and so we can finish the above chain of equalities: $$\mu(\liminf_{n\to\infty} A_n) = \lim_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right) = \liminf_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right) = \liminf_{k\to\infty}a_k \leq \liminf_{k\to\infty}b_k = \liminf_{k\to\infty}\mu(A_k)$$

You're almost there. Note that for every $$k$$ and every $$n\geq k$$ we have $$\bigcap_{i=k}^{\infty}A_i \subset A_n$$, hence, $$\mu\left(\bigcap_{i=k}^{\infty}A_i\right) \leq \mu(A_n)$$. Since this is true for all $$n\geq k$$, we have \begin{align} \mu\left(\bigcap_{i=k}^{\infty}A_i\right) \leq \inf\limits_{n\geq k} \mu(A_n). \end{align} (this inequality is nothing more than the definition of infimum as greatest lower bound). Now, take the limit as $$k\to \infty$$ to get \begin{align} \mu(\liminf_{n\to \infty}A_n)= \lim_{k\to \infty}\mu\left(\bigcap_{i=k}^{\infty}A_i\right) \leq \lim_{k\to \infty}\inf_{n\geq k} \mu(A_n) = \liminf_{n\to \infty}\mu(A_n) \end{align} The final inequality is one of the possible definitions of $$\liminf$$ of a sequence of real numbers (it's simply the limit of the infimum of the "tail" of the sequence).

$$E_n=\bigcap_{m\geq n}A_m$$ is a monotone nondecreasing sequence if measurable sets and $$E_n\subset A_n$$ for all $$n\in\mathbb{N}$$. Hence $$\mu(E_n)\leq \mu(A_n)\qquad\forall n\in\mathbb{N}$$

By definition $$\liminf_nA_n=\bigcup_nE_n$$ and passing to the limit as $$n\rightarrow\infty$$

$$\mu(\liminf_nA_n)=\mu(\bigcup_nE_n)=\lim_n\mu(E_n)=\liminf_n\mu(E_n)\leq\liminf_n\mu(A_n)$$

the first limit follows from the continuity (equivalently $$\sigma$$-additivity property) of a measure. The remaining inequalities follows from elementary properties of numerical limits.

• It may be the case that $$\mu(\liminf_nA_n)=\infty$$: Consider for example $$A_{2n}=[0,\infty)$$ and $$A_{2n+1}=(-\infty,0]$$. Then $$\liminf_mA_n=\mathbb{R}$$. If $$\mu$$ the Lebesgue measure on $$\mathbb{R}$$, then $$\mu(\liminf_nA_n)=\infty=\liminf_n\mu(A_n)$$.

• It may be that $$\mu(\liminf_nA_n)<\infty=\liminf_n\mu(A_n)$$: Consider $$A_n=[n,\infty)$$ and $$\mu$$ the Lebesgue measure on $$\mathbb{R}$$. Then $$\mu(\liminf_nA_n)=0\leq\liminf_n\mu(A_n)=\infty$$.