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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.

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  • $\begingroup$ Have you tried using the fact that $\limsup_{j\rightarrow\infty}f_j(x)=\sup_{i\geq 0}\inf_{j\geq i}f_j(x)$? $\endgroup$ – neth Oct 28 '15 at 0:39
  • $\begingroup$ No, but I have a hard time understanding that, is there way of seeing that result visually? $\endgroup$ – Wolfy Oct 28 '15 at 0:46
  • $\begingroup$ What is your definition of $\limsup$? This is the definition given on page 11 of Folland. $\endgroup$ – neth Oct 28 '15 at 1:03
  • $\begingroup$ ahh I see now thanks its quite simple now $\endgroup$ – Wolfy Oct 28 '15 at 1:24
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    $\begingroup$ @neth Watch out, limsup is an inf of sups, not a sup of infs. $\endgroup$ – Did Oct 29 '15 at 8:17
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@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.

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  • $\begingroup$ Just to clarify we can define $g_{1,k} = \sup_{j\geq k}f_j(x)$ since we have shown that $g_1$ is measurable? To simplify my question I am wondering what gives us the power to set $g_{1,k}$ and $g_{2,k}$ equal to what you set it as. $\endgroup$ – Wolfy Jun 21 '16 at 19:14
  • $\begingroup$ @Wolfy , We started with a sequence $\{f_j\}_{j\geq 1}$. What happens if we throw away the first $k-1$ elements of $\{f_j\}_{j\geq 1}$? We still have a sequence. It is $\{f_j\}_{j\geq k}$. For any sequence of functions we can take the pointwise $\sup$ and $\inf$. So we can define $g_{1,k}$ and $g_{2,k}$ the way we did. Moreover, the function $g_{1,k}$ is just the function $g_1$ built for the sequence $\{f_j\}_{j\geq k}$, so, by Part 1, $g_{1,k}$ is measurable. In a similar way, $g_{2,k}$ is measurable. $\endgroup$ – Ramiro Jun 21 '16 at 19:25
  • $\begingroup$ I see that makes sense, thank you $\endgroup$ – Wolfy Jun 21 '16 at 19:28
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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 $g_4$.

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  • $\begingroup$ Nice, but your proof is pretty complicated $\endgroup$ – Wolfy Oct 28 '15 at 1:16
  • $\begingroup$ Good proof is often not simple. $\endgroup$ – hermes Oct 28 '15 at 4:01
  • $\begingroup$ @MorganWeiss Not sure there is anything simpler (I mean, in the category of correct proofs...). +1. $\endgroup$ – Did Oct 29 '15 at 8:16

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