It sounds like you've solved the problem; if not, I'll help out. For the question concerning Riemann vs. Lebesgue integration, I'll see if I can give you some motivation. Let's just pretend we're working with a smooth nonnegative function $f$ over $\mathbb{R}.$
In Riemann integration, we start by partitioning the $x$-axis, and then we capture the area under the curve by measuring how much of the $y$-axis can fit under the curve above a given element of our partition.
Conversely, with Lebesgue integration, we start to approximate $f$ by approximating the range of $f$ with the $\alpha_i$ used in the definition. In some sense, we're partitioning the $y$-axis into chunks that describe the $y$-behavior of $f$. Then, once we've approximately partitioned the range of $f$, we get a similar notion of area by measuring the $A_i$, which describe the sets of all $x\in \mathbb{R}$ for which the corresponding $\alpha_i$ is a 'good' approximation of $f$. You can see how, in a rough sense, Riemann integration gets area by chopping up $x$ and measuring $y$, and Lebesgue integration gets area by chopping up $y$ and measuring $x$. Only, in the latter case, we have better tools for describing measure.
The standard example is something like $f = \chi_\mathbb{Q}\cap [0,1]$. Obviously, the Riemann upper and lower sums are 1 and 0 respectively, so $f$ is not Riemann integrable, and partitioning the $x$-axis seems unfruitful. On the other hand, if we're using Lebesgue measure, we let $\alpha_0 = 0$ and $\alpha_1 = 1$ and apply your result to get $\int f = 0$. So this $y$-chopping and $x$-measuring lets us handle a wider variety of functions (in general).
The Wikipedia page on Lebesgue integration has a section for motivation/intuition, in case that's helpful.
To show that the Lebesgue integral of $x^{-1}$ is infinite, use a change of variables $x \mapsto 1/x$ and consider the sequence of simple functions
$$\phi_n = \sum_{j=1}^nj \chi_{((j+1)^{-1},j^{-1})}.$$
Note that $0 \leqslant \phi_n(x) \leqslant x^{-1}$ and
$$\int_{(0,1]}\phi_n = \sum_{j=1}^n j\left(\frac1{j}- \frac1{j+1}\right)=\sum_{j=1}^n \frac1{j+1}\to_{n \to \infty} +\infty.$$
Whence,
$$\int_{[1,\infty)}x^{-1} = \int_{(0,1]}x^{-1} = +\infty$$
Best Answer
Since $A_i \cap E \subseteq E$, $\mu(A_i \cap E) \leqslant \mu(E)=0$ by monotonicity.