# Pointwise and uniform convergence

#### 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 $$\displaystyle \{a_n\}$$ be a sequence that contains each rational in [0,1] precisely once. For each $$\displaystyle n\in N$$, define
$$\displaystyle f_n(x)= 0$$ if x is irrational
$$\displaystyle f_n(a_i)=0$$ if $$\displaystyle i > n$$
$$\displaystyle f_n(a_i)=1$$ if $$\displaystyle i \leq n$$
Prove that $$\displaystyle \{f_n\}$$ 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 $$\displaystyle lim f_n=f$$ as n approaches infinity.
I tried this again and I got this far: $$\displaystyle f(x)=0$$ if x is irrational and $$\displaystyle f(x)=1$$ if x is rational.
Consider: absolute value of $$\displaystyle f_n(x)-f(x)$$=$$\displaystyle 0-0=0< \epsilon$$ if x is irrational. Choose $$\displaystyle \epsilon=1/2$$
If x is rational, $$\displaystyle f_n(x)-f(x)=1-1=0< \epsilon$$. I think f(x)=1 because eventually $$\displaystyle i \leq n$$ as n approaches infinity.

2) For $$\displaystyle n \in N$$ define $$\displaystyle f_n: [0,1] \rightarrow R$$ by $$\displaystyle f_n(x)=1/x$$ if $$\displaystyle x \in [1/n,1]$$ and $$\displaystyle f_n(x)=n^2x$$ if $$\displaystyle x \in [0,1/n)$$. Prove that $$\displaystyle \{f_n\}$$ 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 $$\displaystyle f_n$$, so I can't go any anywhere. (Crying) It would be really helpful if someone helps me how to find the pointwise limit

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#### Failure

I got stuck on two problems that involves finding the pointwisse limits of piecewise functions. Hope someone can give me a hand.
1) Let $$\displaystyle \{a_n\}$$ be a sequence that contains each rational in [0,1] precisely once. For each $$\displaystyle n\in N$$, define
$$\displaystyle f_n(x)= 0$$ if x is irrational
$$\displaystyle f_n(a_i)=0$$ if $$\displaystyle i > n$$
$$\displaystyle f_n(a_i)=1$$ if $$\displaystyle i \leq n$$
Prove that $$\displaystyle \{f_n\}$$ 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 $$\displaystyle lim f_n=f$$ as n approaches infinity.
I tried this again and I got this far: $$\displaystyle f(x)=0$$ if x is irrational and $$\displaystyle f(x)=1$$ if x is rational.
Consider: absolute value of $$\displaystyle f_n(x)-f(x)$$=$$\displaystyle 0-0=0< \epsilon$$ if x is irrational. Choose $$\displaystyle \epsilon=1/2$$
If x is rational, $$\displaystyle f_n(x)-f(x)=1-1=0< \epsilon$$. I think f(x)=1 because eventually $$\displaystyle i \leq n$$ 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 $$\displaystyle x\in [0;1]\cap\mathbb{Q}$$, then there exists a unique $$\displaystyle n_0$$ such that $$\displaystyle x=a_{n_0}$$ and therefore for all $$\displaystyle n\geq n_0$$ we have that $$\displaystyle |f_n(x)-1|=|f_n(a_{n_0})-1|=|1-1|=0<\epsilon$$

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

2) For $$\displaystyle n \in N$$ define $$\displaystyle f_n: [0,1] \rightarrow R$$ by $$\displaystyle f_n(x)=1/x$$ if $$\displaystyle x \in [1/n,1]$$ and $$\displaystyle f_n(x)=n^2x$$ if $$\displaystyle x \in [0,1/n)$$. Prove that $$\displaystyle \{f_n\}$$ 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 $$\displaystyle f_n$$, so I can't go any anywhere. (Crying) It would be really helpful if someone helps me how to find the pointwise limit
What about $$\displaystyle f(x) := \begin{cases}0, &\text{if } x=0\\\tfrac{1}{x}, &\text{otherwise}\end{cases}$$

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• jackie