f bounded and measurable implies L-integrable?

Question

I need some help with a the following statement:

Show: Is $f:[0,1] \to \mathbb{R}$ a bounded and measurable function, so f is L-integrable on $[0,1]$

Can I show it with the Lebesgue Theorem (dominated convegence)?

b) Show: If $(f_n)$ is a sequence of bounded measurable functions on $[0,1]$ and convgerts uniformly on $[0,1]$ to $f$, so f is L-integrable on [0,1] and

$\lim int_{[0,1]} f_n d\lambda = \int_{[0,1]} f d\lambda $ ? If my idea for a) is right, cant i take the same proof?

Thanks for helping me

Answer 1

For (a) I would think about the fact that f is bounded and you're integrating on a set of finite measure. For (b) the dominated convergence theorem should do the work for you but try and see how to satisfy the required hypotheses.

Answer 2

For (b) we know f will be bounded since the convergence is uniform (uniform sequence of bounded functions will converge to a bounded function) so $f$ is bounded and measurable. From part (a) we know f is integrable. To show the existence of an integrable function that dominates every term in the sequence you can take $M=sup\{|f_n|:n \in \mathbf{N}, x\in [0,1]\}$ this M will exist and be finite and will dominate all the $f_n$'s at every point in the interval. Now define $g(x)=M \chi_{[0,1]}$ this will be integrable and will allow us to invoke Dominate Convergence.