Is the following function Lebesgue integrable?

Question

Given is the function $$f(x)=\left\{ \begin{array}{rl} \frac{1}{\sqrt{x}}, & x\in \mathbb{I}\cap [0,1]\\ x^3, &x\in \mathbb{Q}\cap [0,1]\end{array}\right.$$

I have to see if $f$ is L-integrable on $[0,1]$. By definition $f$ has to be a measurable bounded function defined on a set $E$ of finite measure to be L-integrable on $E$. I have problems with the boundary part, because if $x=\varepsilon >0$ is a really small irrational number, wouldn't $f(x)\rightarrow \infty$, and thus wouldn't that imply that $f$ isn't bounded.

Answer 1

I will write out an answer in terms of Baby Rudin ch. 11, but be warned there are other ways to construct the Lebesgue integral. Your definition of Lebesgue integrable, which requires that $f$ be bounded, is anomalous. $f$ is obviously not bounded, as you point out.

$\mathbb{Q}\cap [0,1]$ has measure zero because it is countable (Rudin 11.11(f)), and so do any of its subsets for the same reason. The function $f$ is equal to $1/\sqrt{x}$ except on this set of measure zero (assuming $\mathbb{I}=\mathbb{R} \backslash \mathbb{Q}$). For any real $a$, the set $\{x:f(x)>a\}$ equals the set of $x$ for which $1/\sqrt{x}>a$, perhaps plus or minus a countable set of measure zero. The set of $x$ for which $1/\sqrt{x}>a$ is measurable, indeed open. The union and difference of measurable sets are measurable because the measurable sets form a $\sigma$-ring aka $\sigma$-algebra (Rudin 11.10). Thus, $\{x:f(x)>a\}$ is measurable for all $a$, and $f$ is measurable (Rudin 11.13).

The conventional definition of "Lebesgue integrable" for a nonnegative function like $f$ requires only that it be measurable and that its integral be finite (Rudin 11.21-11.22). As for the finiteness of the integral, $\int_0^1 f(x)\,dx$ equals $\int_0^1 dx/\sqrt{x}$ because we can ignore the set of measure zero $\mathbb{Q}\cap[0,1]$ (Rudin 11.25). The latter integral is readily calculated because $d\sqrt{x}/dx=1/(2\sqrt{x})$, and is finite.

Answer 2

$f=f_1 1_{\mathbb{I}\cap [0,1]}+f_2 1_{\mathbb{Q}\cap [0,1]}$, where $f_1(x)=x^{-1/2}$ and $f_2(x)=x^3$. Conclude integrating both sides.

Answer 3

One can write $$f=x\mapsto x^3\mathbb{1}_{\mathbb{Q}\cap[0,1]}(x)+\frac{1}{\sqrt{x}}\mathbb{1}_{\mathbb{I}\cap[0,1]}(x)$$ where I guess that $\mathbb{I}$ is the set of irrational numbers. Now $[0,1]$, $\mathbb{Q}$ and $\mathbb{I}$ are measurable sets, with $\mathbb{Q}$ being negligible. You can conclude that $f$ is measurable.

As for the integral, if $\lambda$ is the Lebesgue's measure, then you have \begin{align}\int_{[0,1]} f\mathrm{d}\lambda & = \int_{[0,1]\cap\mathbb{Q}} f\mathrm{d}\lambda+\int_{[0,1]\cap\mathbb{I}} f\mathrm{d}\lambda \\ & = \int_{[0,1]\cap\mathbb{I}} f\mathrm{d}\lambda \\ & = \int_{[0,1]} f\mathbb{1}_{\mathbb{I}}\mathrm{d}\lambda \\ & = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{x}}. \end{align}