Real Analysis, Folland Proposition 2.7

Question

If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions $$\begin{aligned} g_1(x) = \sup_{j}f_j(x), \ \ \ \ g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x) \end{aligned}$$ $$\begin{aligned} g_2(x) = \inf_{j}f_j(x), \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x) \end{aligned}$$ are all measurable functions. If $f(x) = \lim_{j\rightarrow \infty}f_j(x)$ exists for every $x\in X$, then $f$ is measurable.

proof: $$\begin{aligned} g_1^{-1}((a,\infty)) = \bigcup_{1}^{\infty}f_j^{-1}((a,\infty)), \ \ \ \ \ g_2^{-1}((-\infty,a)) = \bigcup_{1}^{\infty}f_j^{-1}((-\infty,a)) \end{aligned}$$ so $g_1$ and $g_2$ are measurable by proposition 2.3. Now we can define $$\begin{aligned} g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x) = \inf_{k\geq 1}\left( \sup_{j \geq k} f_j(x)\right) \ \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x) = \sup_{k\geq 1}\left(\inf_{j \geq k} f_j(x)\right) \end{aligned}$$ so, $g_3$ and $g_4$ are measurable.

I am not sure if this is right, any suggestions is greatly appreciated.

Answer 1

@Wolfy , Your proof is correct and, in fact, it is a simple and clear proof. I copy it here, just to add some details, to make it even clearer.

If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions $$\begin{aligned} g_1(x) = \sup_{j}f_j(x), \ \ \ \ g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x) \end{aligned}$$ $$\begin{aligned} g_2(x) = \inf_{j}f_j(x), \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x) \end{aligned}$$ are all measurable functions. If $f(x) = \lim_{j\rightarrow \infty}f(x)$ exists for every $x\in X$, then $f$ is measurable.

Proof:

Part 1: $$\begin{aligned} g_1^{-1}((a,\infty)) = \bigcup_{1}^{\infty}f_j^{-1}((a,\infty)), \ \ \ \ \ g_2^{-1}((-\infty,a)) = \bigcup_{1}^{\infty}f_j^{-1}((-\infty,a)) \end{aligned}$$ so $g_1$ and $g_2$ are measurable by proposition 2.3.

Part 2:

Now we can define $g_{1,k}(x)=\sup_{j \geq k} f_j(x)$ and $g_{2,k}(x)=\inf_{j \geq k} f_j(x)$.

For all $k$, we apply Part 1 to the sequence $\{f_j\}_{j\geq k}$, and we have that, $g_{1,k}$ and $g_{2,k}$ are measurable functions.

Now,

$$\begin{aligned} g_3(x) &= \lim_{j\rightarrow \infty}\sup f_j(x) = \inf_{k\geq 1}\left( \sup_{j \geq k} f_j(x)\right)= \inf_{k\geq 1}\left(g_{1,k}(x) \right)\\ g_4(x) &= \lim_{j\rightarrow \infty}\inf f_j(x) = \sup_{k\geq 1}\left(\inf_{j \geq k} f_j(x)\right)=\sup_{k\geq 1}\left( g_{2,k}(x) \right) \end{aligned}$$ so, applying Part 1, to the sequences $\{g_{1,k}\}_k$ and $\{g_{2,k}\}_k$, we conclude that $g_3$ and $g_4$ are measurable.

Answer 2

One has \begin{align*} \limsup f_j(x) \geq a &\iff \inf \sup_{j \geq k} f_j(x) \geq a\\ &\iff \forall k: \quad \sup_{j \geq k} f_j(x) \geq a \\ &\iff \forall k, \forall \epsilon > 0, \exists j \geq k: \quad f_j(x) \geq a - \epsilon\\ &\iff \forall k, \forall n > 0, \exists j : \quad f_j(x) \geq a - \frac{1}{n}\\ &\iff x \in \bigcap_{k=1}^{\infty} \bigcap_{n = 1}^{\infty} \bigcup_{j=k}^{\infty} f_j^{-1}\left(\left[a - \frac{1}{n}, \infty\right)\right) \end{align*} This shows that $$g_3^{-1}([a,\infty)) = \bigcap_{k=1}^{\infty} \bigcap_{n = 1}^{\infty} \bigcup_{j=k}^{\infty} f_j^{-1}\left(\left[a - \frac{1}{n}, \infty\right)\right)$$ Now by assumption, each $f_j^{-1}([a - \frac{1}{n}, \infty))$ is measurable. So are their countable intersections and unions. Reflecting every inequality or working with $-f_j$ will give you the measurability of $g_4$.