I'm studying Lebesgue theory and when a problem asks to actually perform a computation I'm at a loss at what to do. Is the typical way to proceed to observe that an integral is the same as its Riemann integral and then use the standard methods from elementary calculus?
Eg. For a number $\alpha$, define $f(x)= x^{\alpha}$ for $0 < x \leq 1$, and $f(0) = 0$. Compute $\int_0^1 f$.
Do I instead break off subsets of measure zero (the point zero) and compute the rest similarly? Any help is appreciated.
If you want to do partition on the range instead of the domain (Riemann integral), then sum, the formula you can use is: $$ \int^1_0 x^{\alpha}\,dx = \int^{\infty}_0 \mathrm{meas}(\{x\in (0,1] : x^{\alpha} > t\})\,dt.\tag{1} $$ Let's say $-1<\alpha<0 $ (interesting case), let $\epsilon = -\alpha$. For $t\leq 1$: $$ \mathrm{meas}(\{x\in (0,1] : x^{\alpha} > t\}) =\mathrm{meas}\big((0,1]\big)= 1, $$ and for $t>1$: $$\mathrm{meas}(\{x\in (0,1] : x^{\alpha} > t\}) =\mathrm{meas}\big((0,t^{-1/\epsilon})\big) = t^{-1/\epsilon}, $$ therefore the right side of (1) becomes: $$ \int^1_0 x^{\alpha}\,dx = \int^1_0 1\,dt + \int^{\infty}_{1}t^{-1/\epsilon}\,dt = 1 - \frac{1}{-1/\epsilon + 1} = \frac{1}{1+\alpha}. $$ Looks a bit tautological though, unless you wanna do the bookkeeping thing to perform a proper cut off to divide the range of $x^{\alpha}$, and use the simple function to approximate, which will lead to the same formula of (1).