# Prove that $g=\limsup f_n$ and $h=\liminf f_n$ are Lebesgue measurable.

Let $f_n:[0,1]\rightarrow\mathbb{R}$ be a sequence of continuous functions. Prove that $g=\limsup f_n$ and $h=\liminf f_n$ are Lebesgue measurable.

*Given $f_n$ being continuous on a compact set in $\mathbb{R}$, $f_n$ are Riemann integrable, thus, Lebesgue integrable.

Note we can rewrite $g$ and $h$ as the following: $g=\limsup f_n=\bigcap_{n\rightarrow\infty}\bigcup_{k\geq n}f_k$, where $k,n\in\mathbb{N}$. Since countable union/intersection of measurable functions are measurable by the definition of $\sigma$-algebra, $g$ therefore is Lebesgue measurable.

Similarly, $h$ is measurable.

Is the above proof correct? Thank you.

• what do you mean with union/intersection of functions? – Veridian Dynamics Aug 9 '17 at 20:25
• Also, you can use "\limsup" for $\limsup$. Looks better than $limsup$ – Veridian Dynamics Aug 9 '17 at 20:25
• countable unions or countable intersections – 2ndaccount Aug 9 '17 at 20:25
• what does it mean $f\cup g$? it is another function? – Veridian Dynamics Aug 9 '17 at 20:26
• I'm okay with unions and intersections of sets. How do you intersect functions? – Xander Henderson Aug 9 '17 at 20:26

First we show that $g_1:=\sup_k f_k$ and $g_2 :=\inf_k f_k$ are measurable. Note that $$g_1^{-1}(a,\infty] = \bigcup_k f_k^{-1}(a,\infty] \quad\text{and}\quad g_2^{-1}[-\infty,a) = \bigcup_k f_k^{-1}[-\infty,a)$$ are measurable. Therefore the countable supremum and infimum of measurable functions are measurable. Consequently $$\limsup_k f_k = \inf_k(\sup_{j\geq k}f_j) \quad\text{and}\quad \limsup_k f_k = \sup_k(\inf_{j\geq k}f_j)$$ are measurable functions.
We may define $$\limsup_{n\to\infty} a_n := \inf_{n\ge 1} \left( \sup_{k\ge n} a_k \right).$$ The function $\limsup_{n\to\infty} f_n$ is then defined pointwise by $$\left(\limsup_{n\to\infty} f_n\right) (x) := \limsup_{n\to\infty} \left( f_n(x) \right) = \inf_{n\ge 1} \left( \sup_{k\ge n} f_k(x) \right).$$ If we can show that the functions $$g^{\wedge}(x) := \sup_{n\ge 1} g_n(x) \qquad\text{and}\qquad g^{\vee}(x) := \inf_{n\ge 1} g_n(x)$$ are measurable for any sequence $(g_n)$ of measurable functions, then we are done (do you see why?). To do this, we need only show that the preimage of $(a,\infty)$ is measurable for any $a\in\mathbb{R}$. But \begin{align} (g^{\wedge})^{-1}((a,\infty)) &= \bigcup_{n=1}^{\infty} (g_n)^{-1}((a,\infty)). \end{align} But each $g_n$ is measurable, so the union is a countable union of measurable sets, therefore measurable. Hence $g^{\wedge}$ is measurable. By a similar argument (replacing $(a,\infty)$ with $(\infty,a)$, for example), we can show that $g^{\vee}$ is measurable. Now, define $$f_n^{\wedge} := \sup_{k\ge n}f_k.$$ Each $f_n^{\wedge}$ is measurable by the above. Then $$\limsup_{n\to\infty} f_n = \inf_{n\ge 1} \left( \sup_{k\ge n} f_n \right) = \inf_{n\ge 1} f_{n}^{\wedge},$$ which is the infimum of measurable functions, and therefore measurable. The argument, mutatis mutandis, identical for the $\liminf$.