Show that $f(x)=1/x^{1/2}, x \in (0,1]$ is Lebesgue integrable?
Asked
Active
Viewed 520 times
0
-
Continuous functions are Riemann integrable hence Lebesgue integrable. – Chazz Mar 07 '14 at 15:41
-
@Chazz True but this function, being unbounded, is not Riemann integrable. – Did Mar 07 '14 at 15:59
-
If f is bounded defined on a closed, bounded function and it's Riemann integrable then it's Lebesgue integrable over that same interval. – Jawad Mar 07 '14 at 16:05
1 Answers
3
The functions $f_n(x)=\frac1{x^{1/2}}\,1_{(1/n,1]}(x)$ are non-negative, bounded, and measurable. Moreover, $f_n\nearrow f$. By monotone convergence, $$ \int_{(0,1]}f=\lim_n\int_{(0,1]}f_n=\lim_n\int_{1/n}^1\frac{dx}{x^{1/2}}=\lim_n\,2\left(1-\frac1{\sqrt n}\right)=2 $$
Martin Argerami
- 205,756
-
-
-
The $f$ is indeed unbounded. The Lebesgue integral does not require functions to be unbounded. – Martin Argerami Mar 12 '14 at 16:40
-
-
Of course: for each $n$, $0\leq f_n(x)\leq \sqrt n$ for all $x$. – Martin Argerami Mar 12 '14 at 21:45