# Thread: Pointwise and uniform convergence

1. ## Pointwise and uniform convergence

I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
1) Let $\{a_n\}$ be a sequence that contains each rational in [0,1] precisely once. For each $n\in N$, define
$f_n(x)= 0$ if x is irrational
$f_n(a_i)=0$ if $i > n$
$f_n(a_i)=1$ if $i \leq n$
Prove that $\{f_n\}$ converges pointwise on [0,1] to a function that is not R-integrable.
My attempt:
I guess that this sequence of functions converges to the rational characteristic function, but I can't really show $lim f_n=f$ as n approaches infinity.
I tried this again and I got this far: $f(x)=0$ if x is irrational and $f(x)=1$ if x is rational.
Consider: absolute value of $f_n(x)-f(x)$= $0-0=0< \epsilon$ if x is irrational. Choose $\epsilon=1/2$
If x is rational, $f_n(x)-f(x)=1-1=0< \epsilon$. I think f(x)=1 because eventually $i \leq n$ as n approaches infinity.

2) For $n \in N$ define $f_n: [0,1] \rightarrow R$ by $f_n(x)=1/x$ if $x \in [1/n,1]$ and $f_n(x)=n^2x$ if $x \in [0,1/n)$. Prove that $\{f_n\}$ does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

I don't know what the pointwise limit of $f_n$, so I can't go any anywhere. It would be really helpful if someone helps me how to find the pointwise limit

2. Originally Posted by jackie
I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
1) Let $\{a_n\}$ be a sequence that contains each rational in [0,1] precisely once. For each $n\in N$, define
$f_n(x)= 0$ if x is irrational
$f_n(a_i)=0$ if $i > n$
$f_n(a_i)=1$ if $i \leq n$
Prove that $\{f_n\}$ converges pointwise on [0,1] to a function that is not R-integrable.
My attempt:
I guess that this sequence of functions converges to the rational characteristic function, but I can't really show $lim f_n=f$ as n approaches infinity.
I tried this again and I got this far: $f(x)=0$ if x is irrational and $f(x)=1$ if x is rational.
Consider: absolute value of $f_n(x)-f(x)$= $0-0=0< \epsilon$ if x is irrational. Choose $\epsilon=1/2$
If x is rational, $f_n(x)-f(x)=1-1=0< \epsilon$. I think f(x)=1 because eventually $i \leq n$ as n approaches infinity.
The part for x rational sounds a litte sloppy to me, apart from that I don't quite see where your problem is. To fix that sloppiness you should argue that if $x\in [0;1]\cap\mathbb{Q}$, then there exists a unique $n_0$ such that $x=a_{n_0}$ and therefore for all $n\geq n_0$ we have that $|f_n(x)-1|=|f_n(a_{n_0})-1|=|1-1|=0<\epsilon$

f is not integrable because all the lower sums are 0 and all the upper sums are 1.

2) For $n \in N$ define $f_n: [0,1] \rightarrow R$ by $f_n(x)=1/x$ if $x \in [1/n,1]$ and $f_n(x)=n^2x$ if $x \in [0,1/n)$. Prove that $\{f_n\}$ does not converge uniformly on [0,1] but converge uniformly on [M,1] where 0<M<1.

I don't know what the pointwise limit of $f_n$, so I can't go any anywhere. It would be really helpful if someone helps me how to find the pointwise limit
What about $f(x) := \begin{cases}0, &\text{if } x=0\\\tfrac{1}{x}, &\text{otherwise}\end{cases}$