0

and $(f_n)_n$ a sequence of measurable function from $\mathbb{R}$ to $\mathbb{R}$ (where $\mathbb{R}$ is equipped with the Borel's $\sigma$-algebra).

Prove that the following two sets are measurable $$A=\{x \in \mathbb{R} \ | \lim_{n\to \infty} f_n(x)=+\infty \}, \qquad B=\{x\in \mathbb{R}\ |(f_n(x))_n\text{ is bounded}\}.$$

  • Not that it changes this particular question, but it is generally dangerous to assume that there is a "usual" sigma-algebra on the reals. – Justpassingby Jan 14 '16 at 15:05
  • What are your own ideas or attempts? – Rhjg Jan 14 '16 at 15:05
  • There must be something missing in the definition of $A.$ Do you mean those $x$ where the limit exists as a real number? – Justpassingby Jan 14 '16 at 15:06
  • I think you are missing something in your definition of $A$. – siegehalver Jan 14 '16 at 15:06
  • The domain of the $f_n$ is said to be the measurable space $E$, yet the argument in the displayed formulas for $A$ and $B$ is in $\mathbb{R}$. Are we to assume that $E\subset \mathbb{R}$? – ForgotALot Jan 14 '16 at 15:12
  • @ForgotALot , sorry guys, I corrected it now. –  Jan 14 '16 at 15:49
  • 1
    Set $A$ is - in spite of your edit - still not well defined. Do you mean $A:={x\in E\mid (f_n(x))_n\text{ is converging}}$? – drhab Jan 14 '16 at 15:58
  • Same comment after your second edit. If set $A$ is defined as I suggested in my former comment then have a look at this answer. – drhab Jan 14 '16 at 16:20
  • @drhab can u take another look now ? –  Jan 16 '16 at 15:22
  • @Justpassingby from what I have experienced so far it seems to be very common in the literature not to mention that $\mathbb R$ is equipped with the Borel $\sigma$ algebra when talking about measurable real valued functions. – noctusraid Jan 16 '16 at 17:28

2 Answers2

1

Write $f(x)=inf f_{n}(x)$. Note that $$\{x\in X:f(x)\leq\alpha\}=\bigcap_{n=1}^{\infty}\{x\in X:f_{n}(x)\leq\alpha\}$$ Thus $f$ is measurable. Similarly, $sup f_{n}(x)$ is measurable. It follows that $$\lim_{n\to\infty}f_{n}(x)=\lim\ inf (f_{n}(x))=sup_{n\geq1}\{inf_{m\geq n} f_{m}(x)\}$$ is measurable.

Rafael
  • 3,975
0

Let $\mathbb N$ denote the set of positive integers.

The following statements are equivalent:

  • $\lim_{n\to\infty}f_n\left(x\right)=+\infty$

  • $\forall m\in\mathbb{N}\exists k\in\mathbb{N}\forall n\geq k\; f_{n}\left(x\right)\geq m$

  • $x\in\bigcap_{n=1}^{\infty}\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\geq m\right\} $

This tells us that: $$A:=\left\{ x\in\mathbb{R}\mid \lim_{n\to\infty}f_n\left(x\right)=+\infty\right\} =\bigcap_{n=1}^{\infty}\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\geq m\right\} $$

The RHS is measurable because for each $n\in\mathbb{N}$ the set $\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\geq m\right\} $ is measurable.

The following statements are equivalent:

  • $\left(f_{n}\left(x\right)\right)_{n}$ is bounded

  • $\exists m\in\mathbb{N}\forall n\in\mathbb{N}\; f_{n}\left(x\right)\leq m$

  • $x\in\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\leq m\right\} $

This tells us that: $$B:=\left\{ x\in\mathbb{R}\mid\left(f_{n}\left(x\right)\right)_{n}\text{ is bounded}\right\} =\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\leq m\right\} $$

The RHS is measurable because for each $n\in\mathbb{N}$ the set $\left\{ y\in\mathbb{R}\mid f_{n}\left(y\right)\leq m\right\} $ is measurable.

drhab
  • 153,781