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?

  • $\begingroup$ You're almost there. Note that $\bigcap_{n=k}^\infty A_n\subseteq A_k$. $\endgroup$
    – Jakobian
    Commented Sep 23, 2020 at 22:25
  • 2
    $\begingroup$ You can see it as Fatou's lemma applied to indicator functions $1_{A_n}$. $\endgroup$
    – user239203
    Commented Sep 23, 2020 at 22:35
  • $\begingroup$ @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... $\endgroup$
    – JoZ
    Commented Sep 23, 2020 at 22:51
  • $\begingroup$ @JoZ it doesn't really make sense. But you can take this inequality and put $\liminf$ on both sides. What do you get? $\endgroup$
    – Jakobian
    Commented Sep 23, 2020 at 22:55

3 Answers 3


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 $n\rightarrow\infty$


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 $\mu(\liminf_nA_n)=0\leq\liminf_n\mu(A_n)=\infty$.


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