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This problem comes from Royden's Real Analysis, 4th ed., pg 84, #19:

For a number $\alpha$, define $f(x)=x^\alpha$ for $0<x\le 1$ and $f(0)=0$. Compute $\int_0^1 f$.

MY WORK:

I know that if $\alpha\ge 0$, then $f(x)=x^\alpha$ is a bounded function on $[0,1]$ and is Riemann integrable. Thus it is Lebesgue integrable and the two integrals are equal. So when $\alpha\ge 0$ we have $$ \int_0^1 x^\alpha=\frac{x^{\alpha+1}}{\alpha+1}\Bigg\vert_0^1=\frac{1}{\alpha+1}. $$ For the case when $-1<\alpha<0$, I want to define an increasing sequence of nonnegative measurable functions on $[0,1]$. So let $$ f_n=\begin{cases}x^\alpha&x\in[1/n,1]\\0&x\in[0,1/n)\end{cases} $$ and I want to consider the sequence of integrals $\{\int_0^1 f_n\}$. For each $n$ we have $f_n(x)$ is a bounded function on $[1/n,1]$ and is Riemann integrable. Thus for each $n$ it is Lebesgue integrable and the two integrals are equal. So $$ \int_0^1f_n=\int_0^{\frac{1}{n}}f_n+\int_{\frac{1}{n}}^1f_n=\int_0^{\frac{1}{n}}0+\int_{\frac{1}{n}}^1x^\alpha=\frac{x^{\alpha+1}}{\alpha+1}\Bigg\vert_{\frac{1}{n}}^1=\frac{1}{\alpha+1}-\frac{\left(\frac{1}{n}\right)^{\alpha+1}}{\alpha+1}\le\frac{1}{\alpha+1}. $$ Thus the sequence $\{\int_0^1 f_n\}$ is bounded by $\frac{1}{\alpha+1}$ for all $n$. Then by Beppo Levi's Lemma, $f_n$ converges pointwise to the measurable function
$$ f(x)=\begin{cases}x^\alpha&x\in(0,1]\\0&x=0\end{cases} $$ and $$ \int_0^1 f=\lim_{n\to\infty}\int_0^1 f_n=\lim_{n\to\infty}\frac{1}{\alpha+1}-\frac{\left(\frac{1}{n}\right)^{\alpha+1}}{\alpha+1}=\frac{1}{\alpha+1}. $$

QUESTION:

I am not sure what to do for the case when $\alpha\le -1$. Any ideas? I am certain that it must diverge to $\infty$, but I am not sure how to use the tools at my disposal to verify this.

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For $x \in [0,1]$ and $\alpha \leqslant -1 $ we have $x^{\alpha} \geqslant x^{-1}$.

Hence,

$$\int_{[0,1]}x^{\alpha}\geqslant \int_{[0,1]}x^{-1}= \infty.$$

To show that the Lebesgue integral of $x^{-1}$ is infinite, 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

$$\int_{(0,1]}\phi_n = \sum_{j=1}^n j\left(\frac1{j}- \frac1{j+1}\right)=\sum_{j=1}^n \frac1{j+1}\to_{n \to \infty} \infty$$

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    $\begingroup$ @Laars Helenius: You're welcome. You could simply show the inproper Riemann integral diverges, but I assumed you have to work with the Lebesgue integral. $\endgroup$ – RRL Jan 23 '15 at 4:07
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Recall in Calculus:

For $\alpha\leq -1$

$\int_0^1x^{-5} = lim_{n-->0}$$ \int_0^1x^{-5}=$$ lim_{n-->0}$ $-\frac{1}{4x^4}|_0^1=$ $-\frac{1}{4}lim_{n-->0}$ $(1-\frac{1}{n^4})=\infty$

I honestly think that this should be enough for $\alpha\leq -1$. This is Riemann integrable, right?

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  • $\begingroup$ i also would like to ask how the simple function has come about? why that definition @RRL $\endgroup$ – desperatemuch Oct 14 '15 at 13:24

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