The limit superior and limit inferior

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The notion of a limit plays a very big role in measure and integration theory. The way how limits interact with integrals and how they behave under certain situations (for example changing the order of taking limits and integrating) lead to the famous limit theorems of Lebesgue integration. We need to recall the notion of the limsup and the one for the liminf in order to get a better intuition of how sequences of measurable function behave. Let therefore [math](a_n)_{n\in\mathbb{N}}[/math] be a sequence in [math]\bar \R[/math] and define the limit superior and the limit inferior of [math](a_n)[/math] as

[[math]] \limsup_{n\to\infty} a_n=\lim_{n\rightarrow\infty}\swarrow\left(\sup_{k\geq n} a_k\right)=\inf_n \sup_{k\geq n}a_k [[/math]]

[[math]] \liminf_{n\to\infty} a_n=\lim_{n\rightarrow\infty}\nearrow\left(\inf_{k\geq n} a_k\right)=\sup_n\inf_{k\geq n}a_k [[/math]]

Note the crucial thing that the above limits always exists in [math]\bar \R[/math].

Proposition

Let [math](E,\A)[/math] be a measurable space and let [math](f_n)_{n\in\N}[/math] be a sequence of measurable maps such that [math]f_n:E\longrightarrow \bar \R[/math]. Then

• [math]\sup_n f_n[/math]
• [math]\inf_n f_n[/math]
• [math]\limsup_n f_n[/math]
• [math]\liminf_n f_n[/math]

are measurable. Im particular, if [math]f_n\xrightarrow{n\to\infty} f[/math] then [math]f[/math] is also measurable. In general we can say that [math]\{x\in E\mid \lim_{n\to\infty}f_n(x)\text{ exists}\}[/math] is measurable.

Show Proof

Let us first define [math]f(x):=\inf_n f_n(x)[/math]. Now we see that it is enough to show that for all [math]a\in\R[/math] we have [math]f^{-1}([-\infty,a))\in\A[/math]. Indeed, we can observe that

[[math]] f^{-1}([-\infty,a))=\left\{ x\in E\mid \inf_n f_n(x) \lt a\right\}=\bigcup_n\left\{ x\in E\mid f_n(x) \lt a\right\}\in\A, [[/math]]

and therefore we can say that [math]f^{-1}([-\infty,a))=\bigcup_n\left\{ f_n \lt a\right\}[/math]. Moreover, we have that

[[math]] \left\{ x\in E\mid \lim_{n\to\infty} f_n(x) \text{ exists}\right\}=\left\{ x\in E\mid \liminf_n f_n(x)=\limsup_n f_n(x)\right\} =G^{-1}(\Delta)\in \A, [[/math]]

where [math]G[/math] is the map given by [math]G:x\longmapsto (\liminf_n f_n(x),\limsup_n f_n(x))[/math] and [math]\Delta[/math] is the diagonal of [math]\bar \R^2[/math], which is closed and hence measurable.

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Example

If a map [math]f:\R\longrightarrow \R[/math] is differentiable, its derivative [math]f'[/math] will be measurable and we can hence write it as a limit of measurable functions as

[[math]] f'(x)=\lim_{n\to\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right). [[/math]]

Let [math](E,\A)[/math] be a measurable space and let [math](f_n)_{n\in\N}[/math] be a sequence of measurable functions [math]f_n:E\longrightarrow X[/math] to some space [math]X[/math]. If [math]\bar \R[/math] is described as the metric space [math](X,d)[/math], we get that [math]f_n\xrightarrow{n\to\infty}f[/math] implies that [math]f[/math] is measurable. Moreover, [math]f_n\xrightarrow{n\to\infty}f[/math] if and only if for all [math]x\in E[/math] we get [math]\lim_{n\to\infty}f_n(x)=f(x)[/math] if and only if for all [math]x\in E[/math] we get [math]\lim_{n\to\infty}d(f_n(x),f(x))=0[/math]. If we consider a closed set [math]F\subset X[/math], we get

[[math]] \begin{align*} f^{-1}(F)&=\left\{x\in E\mid d(f(x),F)=0\right\}=\left\{ x\in E\mid \lim_{n\to\infty}d(f_n(x),F)=0\right\}\\ &=\left\{x\in E\mid \forall p\geq 1,\exists N\in\N \text{ such that $\forall n\geq N$ } d(f_n(x),F)\leq \frac{1}{p}\right\}\\ &=\bigcap_{p\geq 1}\bigcup_{N\in\N}\bigcap_{n\geq N}\left\{ x\in E\mid d(f_n(x),F)\geq \frac{1}{p}\right\}\in\A. \end{align*} [[/math]]

If we consider a complete metric space [math](E,d)[/math], then one can show that [math]\{x\in E\mid f_n(x) \text{ converges}\}\in \A[/math]. We leave this as an exercise for the reader.

Definition (Push-forward measure)

Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measurable spaces and let [math]\mu[/math] be a positive measure on [math](E,\A)[/math]. Moreover, let [math]f:E\longrightarrow F[/math] be a measurable map. Then the push-forward of the measure [math]\mu[/math] by [math]f[/math], denoted by [math]f_*\mu[/math] is defined for all [math]B\in\B[/math] as

[[math]] f_*\mu(B):=\mu(f^{-1}(B)). [[/math]]

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].