I am considering the measure space $([0,1),\mathcal{B}([0,1),\lambda)$ and for every whole number $n\geq1$ I have the function:
$u_{n}=nx^{n-1}-(n+1)x^{n}$ , $x \in [0,1)$
I am asked the following:
a) show that $u_{n} \in \mathcal{L}^{1}([0,1),\lambda))$
b) find $\sum_{n=1}^{\infty}\int_{[0,1)}u_{n}(x)d\lambda(x)$
I have done the following (referring to the theorems in my book):
a) Since $u_{n}$ is continuous then it is measurable and Riemann integrable. Since it is measurable Riemann integrable then it is Lebesgue integrable and therefore $u_{n} \in \mathcal{L}^{1}([0,1),\lambda))$ and we have that $\int_{[a,b]}ud\lambda =(R)\int_{a}^{b}u(x)dx$.
b) Using the last part of a) we have:
$\sum_{n=1}^{\infty}\int_{[0,1)}u_{n}(x)d\lambda(x)= \\ \sum_{n=1}^{\infty}(R)\int_{0}^{1} (nx^{n-1}-(n+1)x^{n})dx \\ \sum_{n=1}^{\infty}0 = 0$
Are a) and b) correct?
Thanks.
You have to be more carefull.
Riemman integrability is certainly valid for continuous functions on closed bounded intervals.
But there are continuous functions on $[0,1)$ that are not Riemman integrable, like $f(x)=\frac{1}{1-x}$
But in this case the functions Riemman integrable since they are bounded and you can extend them continuously to $1$
So the Riemman integral 0f $_n$ will be equal to the Lebesgue integral
The rest is correct.