I'm learning about measure theory, specifically Lebesgue integration, and need help to understand the solution to the following problem:
Let $f_n(x) = nx^{n-1}-(n+1)x^n$, $x\in (0, 1)$. Show that
$$\int_{(0, 1)}\sum_{n=1}^{\infty}f_n \neq \sum_{n=1}^{\infty}\int_{(0, 1)}f_n.$$
Solution. Obviously $\underbrace{\sum_{n=1}^{\infty}f_n(x) = 1}_{(1)}$ and therefore $\int_{(0, 1)}\sum_{n=1}^{\infty}f_n=1$.
On the other hand, $\underbrace{\int_{(0, 1)}f_n=0}_{(2)}$, $n=1, 2, \cdots$.
It is not obvious to me that the sum $(1)$ is equal to one. Neither is it obvious that the Lebesgue integral $(2)$ is equal to zero for every $n \in \mathbb{N}$. I'm looking for an explanation that would help me understand both $(1)$ and $(2)$.
Edit: Considering the hint of Zachary Selk for $(2)$, we have
$$\begin{align} \int_{(0,1)}f_n &= \int_{(0,1)}nx^{n-1}-(n+1)x^n\\ &= \int_{(0,1)}nx^{n-1}-\int_{(0,1)}(n+1)x^n\\ &= \left[\frac{nx^n}{n}\right]_{0}^{1}-\left[\frac{(n+1)x^{n+1}}{n+1}\right]_{0}^{1} \\ &= (1^n-0^n)-(1^{n+1}-0^{n+1})\\ &= 1-1\\ &= 0. \end{align}$$
For 1, just note that it's a telescoping sum. That is:
$$\sum_{n=1}^k (nx^{n-1}-(n+1)x^n) =(1x^{1-1}-(1+1)x^1)+\cdots+(kx^{k-1}-(k+1)x^k)=1-(k+1)x^k$$
Just take the limit as $k \to \infty$ and you get $1$
For 2, just do the integral, The antiderivative is just the power rule.