Showing that $f_n \to f$ in $L^1{[0,1]}$ Suppose  
(a) $f_n \to f$ a.e. on $[0,1]$ and
(b) $\|f_n\|_{L^\infty_{(0,1)}} \leq M \lt \infty.$  
I want to show that $f$ is Lebesgue integrable and $f_n \to f$ in $L^1{[0,1]}$.     
Also, I want an example where the condition (b) is not satisfied and the conclusion fails.   

 A: Part (a) follows directly from the Dominated Convergence Theorem since the constant function $M$ is in $L^1$. For the second part define $f_n(x)$ to be zero on the interval $[1/n,1]$ and make the function large enough so that $\int_0^{1/n} f_n(x)dx=1$.
A: That $f_n\to f$ in $L^1$ follows from Lebesgue's dominated convergence theorem.
For the counterexample, let $f_n(x)=n\ \chi_{[0,1/n]} (x)$, where:
$$\chi_{[0,1/n]} (x):=\begin{cases} 1 &\text{, if } 0\leq x\leq 1/n \\ 0 &\text{, otherwise.}\end{cases}$$
Then $f_n\to 0=:f$ a.e. in $[0,1]$ but $\lVert f_n\rVert_\infty\to \infty$ and:
$$\int_0^1 |f_n| = 1 \qquad \text{and} \qquad \int_0^1 |f| =0\; ,$$
therefore $f_n\not\to f$ in $L^1$.
A: Part (a) follows directly from Lebesgue's Dominated Convergence Theorem, where the dominating function is the constant function equal to $M$.
http://en.wikipedia.org/wiki/Dominated_convergence_theorem
For part (b), consider a function that is 0 everywhere outside $[1/n,1]$ but which "spikes" at $\frac{1}{2n}$ to a height of $2n$. (ie, let $f(x)=4n^2x$ when $x\leq 1/2n$ and $f(x)=4n^2(1/n-x)$ when $1/2n\leq x\leq 1/n$.) Then $\int f_n=1$ for all $n$ but $\int f=0$.
