1. Uniform convergence in measures

Just stuck on a Measure theory question right now:

Suppose $\displaystyle \{f_n\}\subset {L^1} and {f_n}\longrightarrow f$ uniformly. Show that if $\displaystyle \mu(X)<\infty , then f\in {L^1}$ and $\displaystyle \int f_n\longrightarrow \int f$. Find an example that shows that the result is not necessarily true when $\displaystyle \mu(X)= \infty$

Little bit stumped on how to start. Initially I thought I could use the dominated convergence theorem, but not fully sure. We do have uniform convergence and the dominated convergence theorem claims if we can bound a function by g then the $\displaystyle \int f =lim_{n\rightarrow\infty} \int {f_n}$.

If not I'm thinking of using chebychev's inequality.

Any help would be great.

I also apologise for any problems with the LaTex, it's my first time using it.

Thanks

2. Since $\displaystyle f_n \rightarrow f$ uniformly we have that for all $\displaystyle x\in X$ there exists an $\displaystyle N$ such that $\displaystyle m,n\geq N$ implies $\displaystyle \vert f_n(x)\vert - \vert f_m(x) \vert \leq \vert f_m(x)-f_n(x) \vert \leq 1$ now taking $\displaystyle m=N$ we get $\displaystyle \vert f_n(x) \vert \leq 1+ \vert f_N(x)\vert$ for all $\displaystyle n\geq N$ and this last one is integrable since $\displaystyle \int_{X} d\mu =\mu (X)<\infty$ so picking $\displaystyle g(x)=\max \{ \vert f_1(x) \vert ,...,\vert f_{N-1}(x) \vert , 1+ \vert f_N(x) \vert \}$ we get our dominating function.

When $\displaystyle \mu (X)= \infty$ see here

3. uniform convergence

looks like what u used cauchy sequence rather than uniformly convergence. can you still do what you did with uniform convergent sequence?

4. $\displaystyle |\int_Xf_n-f|\le\int_X|f_n-f|\le\mu(X)\cdot\sup|f_n-f|<\epsilon\mu(X)$.

For a counter example consider $\displaystyle f_n=\frac{1}{n}\chi_{[0,n]}$. Then $\displaystyle f_n\to 0$ but $\displaystyle \int_\mathbb{R}f_n=1$ for all $\displaystyle n$.

5. L1 space

can you show me how to show that $\displaystyle f$ is in L1, please?

6. I think that the following version of the triangle inequality should help: $\displaystyle |a|-|b|\le |a-b|$