Counterexample for convergence in L1

**tttcomrader** · November 24th 2009, 08:14 AM

Suppose that $\{ f_n \} \subset L^1( \mu )$ and $f_n$ converges uniformly to f [/tex], by a theorem that I know, $f \in L^1 ( \mu )$ and $\int f_n \rightarrow \int f$ provided that $\mu (X) < \infty$ , X being the domain set.

Now, if $\mu (X) = \infty$ , this may not be true.

Counterexample:

Define $f_n(x)= \left \{ \begin {array} {cc} \frac {1}{x} & \mbox { if }<br /> 1 \leq x < n \\ 0 & \mbox { if } x \geq n \end {array} \right .$

And define $f(x)= \left \{ \begin {array} {cc} \frac {1}{x} & \mbox { if }<br /> 1 \leq x \\ 0 & \mbox { if } x \leq 1 \end {array} \right .$

And $f_n \rightarrow f$ uniformly, but $f_n$ is integrable for all n while f is not.

Question:

Q1: Why is $f_n \rightarrow f$ uniformly? The book just claim it is, but I'm trying to prove it using the defintion of $\forall \epsilon > 0 \exists N \in \mathbb {N}$ such that $\mid f_n-f \mid < \epsilon$

Q2: I understand why f is not integrable since it goes to infinity, but are $f_n$ integrable? Is it because $\int f_n = \int _1^n \frac {1}{x}$ is finite?

Thank you!

**Jose27** · November 24th 2009, 11:04 AM

1) I assume your domain is $\mathbb{R}$ notice that to evaluate $\vert f_n(x)-f(x) \vert$ we need three cases:

-) If $x< 1$ then $f_n(x)=f(x)=0$
-) If $1\leq x \leq n$ then $f_n(x)=f(x)=\frac{1}{x}$
-) If $n<x$ then $\frac{1}{x}$ $\leq \frac{1}{n}$ and $\vert f_n(x) -f(x) \vert = \frac{1}{x} < \epsilon$ for $n$ big enough.

2) Notice that $f_n$ are Riemann integrable.

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