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Let $(\Omega ,A,\mu )$ be a measure space and $f,\,f_1,\ldots,\,f_n,\ldots$ measurable functions such that: $(1)$ $f_n \overset{\mu }{\longrightarrow}f$ and $(2)$ $|f_n(x)|\leq 1$ for a.e. $x$ for all $n$. Show that $|f(x)| \leq 1$ for a.e. $x$.

What I have done so far:

$f_n \overset{\mu }{\rightarrow}f \Leftrightarrow \forall \varepsilon >0:\mu (\left \{ x:|f_n(x)-f(x)|\geq \varepsilon \right \})\xrightarrow[]{n\rightarrow \infty }0$

$\forall n,\exists N \in \Omega : \mu (N)=0 \wedge\forall x \in \Omega \setminus N : |f_n(x)|\leq 1$

$|f_n (x)-f(x)|\leq |f_n(x)|-|f(x)| \leq 1 - |f(x)| $

But now I have no idea how I should continue...

Clayton
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Bünzli Refinej
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    $|f_n(x)-f(x)| \leq |f_n(x)|-|f(x)|$ is not true. – JPMarciano Nov 04 '21 at 15:39
  • I've edited the post to clean up the presentation a bit. I tried as best as I could to maintain the OP's original intent; feel free to roll back the edit if it seems more unclear or if I entered something incorrectly. – Clayton Nov 04 '21 at 15:43
  • Suggestion: Write down what it is that you want to prove. Compare this to the information that the problem gives to you and see if there are ways to implement the given information into the statement you want to prove. – Clayton Nov 04 '21 at 15:45

2 Answers2

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Study the set $$\mu (\left \{ x:|f(x)|\geq 1 \right \}).$$ How $|f| \leq |f_n-f|+|f_n|$ it follows that $$ \left \{ x:|f(x)|\geq 1+2\varepsilon \right \} \subset \{ x:|f_n(x)-f(x)|\geq \varepsilon \} \cup \{ x:|f_n(x)|> 1 \}.$$ for small $\varepsilon$.Then $$ \mu \left( \{ x:|f(x)|\geq 1+2\varepsilon \right \}) \leq \mu \left( \{ x:|f_n(x)-f(x)|\geq \varepsilon \} \right) + \mu \left( \{ x:|f_n(x)|>1 \} \right).$$ Now send $n$ to infinity and $\varepsilon \rightarrow 0$ to get the result.

Ilovemath
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  • Did you mean ${x:|f(x)|\geq 1}\subset{x:|f_n(x)-f(x)|\geq \varepsilon}\cup {x:|f_n(x)|\geq 1-\varepsilon}$? It doesn't seem right either way. – JPMarciano Nov 04 '21 at 15:57
  • I added an edit and fixed it. – Ilovemath Nov 04 '21 at 16:03
  • If we send $n$ to infinity, $\mu({x:|f_n(x)|\geq 1-\varepsilon})$ doesn't necessarily go to zero, since it can be the case that $f_n(x)=1$ for every $x \in \Omega$ for example. – JPMarciano Nov 04 '21 at 16:06
  • $\varepsilon$ is arbitrary by definition of measure convergence, it can be $1/n.$ – Ilovemath Nov 04 '21 at 16:13
  • so if we send n to infinity the right one goes to 0 , but for the left one we have $\mu \left( { x:|1-f(x)|\geq \varepsilon } \right)$ is this smaller than $\mu \left( { x:|1| +|-f(x)|\geq \varepsilon } \right)$ ? i'm sorry this is all very new for me and i don't really know what i am allowed to do here – Bünzli Refinej Nov 04 '21 at 16:35
  • when you talk on the left you are talking about $\mu \left( { x:|f_n(x)-f(x)|\geq \varepsilon } \right)$? If so, it's because $f_n \overset{\mu }{\rightarrow}f$. – Ilovemath Nov 04 '21 at 16:40
  • ah ok, so $\mu \left( { x:|f_n(x)-f(x)|\geq \varepsilon } \right)$ is also going to 0 ? – Bünzli Refinej Nov 04 '21 at 16:43
  • Yes! exactly because $f_n$ converges to $f$ in measure. – Ilovemath Nov 04 '21 at 16:44
  • aaaah ok and then ${x:|f(x)|\geq 1}\subset 0 \cup 0$ ? and this shows that $|f (x) | \geq 1$ ? a.e i forgot – Bünzli Refinej Nov 04 '21 at 16:46
  • no, you are confusing set inclusion with sum of measures. You have to pay more attention. If you are having difficulty understanding convergence in measure, I recommend reading Bartle's book: https://books.google.com.br/books?id=aE1YBAAAQBAJ&printsec=frontcover&redir_esc=y#v=onepage&q&f=false for example, see chapter 7. – Ilovemath Nov 04 '21 at 17:04
  • @Ilovemath even if $\varepsilon=\frac{1}{n}$ it can be the case this term doesn't go to 0. For example, if $f_n(x)=1$ for every $n \in \mathbb{N}$ and $ x \in \Omega$ then the term is always equal to $\mu(\Omega)$ for every $\varepsilon>0$ and $n \in\mathbb{N}$. – JPMarciano Nov 04 '21 at 17:36
  • @JPMarciano a new edit was made, now it's clearer. – Ilovemath Nov 04 '21 at 18:34
  • @Ilovemath It doesn't seem right yet for the same reason: $\mu({x:f_n(x)\geq 1})=\mu(\Omega)$ when $f_n(x)=1$ for every $x$ and $n$. – JPMarciano Nov 04 '21 at 21:20
  • I think it's just a matter of modifying the inclusion of sets as above. – Ilovemath Nov 04 '21 at 21:36
  • Yeah, now it is ok! Sorry for being annoying. – JPMarciano Nov 04 '21 at 21:40
  • no problem! math needs to be clear and correct! – Ilovemath Nov 04 '21 at 22:01
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I would try using the following inequality: $$|f| \leq |f_n| + |f - f_n|.$$

Gary Moon
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