Evaluate the limit of series, Lebesgue integral

Question

I am about to evaluate the Lebesgue integral $\int_{[0,1]}x^d dm$, for $d \ge 1$, without using Riemann Integral

My thinking is to for any $n \ge 1$, partition the range into $I_{n,i} = [\frac{i}{n},\frac{i+1}{n}]$, $i = 0,\ldots,n-1$

Then define the simple function: $\phi_n = \sum_{i=0}^{n}\frac{i}{n} \chi_{[(\frac{i}{n})^{1/d}, (\frac{i+1}{n})^{1/d}],}$, then $\phi_n \to x^d$ is increasing, using ${\bf MCT}:$

$\int_{[0,1]} x^d dm = \lim_{n \to \infty}\sum_{i=0}^{n} \frac{i}{n}[(\frac{i+1}{n})^{1/d} - (\frac{i}{n})^{1/d}]$

So my question is how to calculate the limit LHS, I will appreciate any help :)

Answer 1

For your approximating function, it shouldn't be hard to see that (note the sum should have $n$ terms, so not $\sum_{i=0}^n\dots$) $$\phi_n(x) = \sum_{i=0}^{n-1} \frac{i}{n} \chi_\left[(\frac in)^{1/d}< x <(\frac {i+1}n)^{1/d}\right] =\sum_{i=0}^{n-1} \frac{1}{n} \chi_\left[(\frac in)^{1/d}< x \right]$$ so that \begin{align}\int \phi_n \, \text dm &= \sum_{i=0}^{n-1} \frac{i}{n} m\left[x:\left(\frac in\right)^{1/d}< x <\left(\frac {i+1}n\right)^{1/d}\right]\\ &=\sum_{i=0}^{n-1} \frac{1}{n} m\left[x:\left(\frac in\right)^{1/d}< x \right] \\ &=\sum_{i=0}^{n-1} \frac{1}{n} \left(1 - \left(\frac in\right)^{1/d}\right) \end{align} and this last sum is a Riemann sum for $\int_0^1 1-t^{1/d} \, \text dt = \frac1{d+1}$.

You did say you don't want to use Riemann integration, but one formulation of the Lebesgue integral is as a Riemann integral of the decreasing rearrangement.

Answer 2

I think it would be a lot simpler to partition the range into intervals of unequal size. If you let $$I_{n,i} = \left[ \left(\frac in \right)^d, \left( \frac {i+1}n \right)^d \right)$$ notice that the length of each $I_{n,i}$ can be estimated by the mean-value theorem: there exists $y \in I_{n,i}$ satisfying $$ \left( \frac {i+1}n \right)^d - \left(\frac in \right)^d = dy^{d-1} \frac 1n \le \frac dn.$$

Let $f(x) = x^d$ and let $E_{n,i} = \{x \in [0,1] : f(x) \in I_{n,i} \}$ then the function $$\phi_n = \sum_{i=0}^{n-1} \left( \frac in \right)^d m(E_{n,i})$$ satisfies $0 \le f - \phi_n \le \frac dn$ so that $$ \int_{[0,1]} \phi_n \to \int_{[0,1]} f$$ using e.g. LDCT. Unlike above, here $E_{n,i} = [i/n,(i+1)/n)$ so that $m(E_{n,i}) = \frac 1n$ and $$ \int_{[0,1]} \phi_n = \sum_{i=0}^{n-1} \frac{i^d}{n^{d+1}}.$$

This limit is simple to evaluate using the well-known formula $$\sum_{i=1}^n i^d = \frac{n^{d+1}}{d+1} + P_d(n)$$ where $P_d$ is a polynomial of degree (at most) $d$.