Why is the Riemann $\int_0^1 1 / \sqrt{x}$ not defined?

Question

On $(0, 1]$, let $f(x) = 1/\sqrt{x}$ with $f(0) = 0$. Let $I = [0,1]$.

Is it then true that the Lebesgue $\int_I f < \infty$, while the Riemann $\int_0^1 f$ isn't even defined? If so, why is the Riemann $\int_0^1 f$ not defined?

Answer 1

By definition the proper Riemann integral acts on bounded functions. It is also only defined on compact intervals, though that is not really the problem here (one can define $f(x)=x^{-1/2}$ on $(0,1]$ and $0$ at $x=0$ for example).

The restriction to bounded functions is motivated by the simple fact that if you have an unbounded function then you can choose your partition to "zoom in" on the singularity which will hinder convergence. Assuming you did the change I mentioned above, you could see the problem by considering the partition $\{ 0,1/n,\dots,1 \}$ and then choosing the evaluation point in $[0,1/n]$ to be, say, $n^{-4}$. Then there is no hope of convergence because the Riemann sums for this sequence of partitions and evaluation points exceed $n$.

This provides an interesting example of a failure of interchange of limits: if we define

$$g : \mathbb{N} \times \mathbb{N} \to \mathbb{R},g(m,n)=\sum_{k=0}^{n-1} \left ( \frac{1}{m} + \left ( 1-\frac{1}{m} \right ) \frac{k}{n} \right )^{-1/2} (1/n)$$

then $\lim_{n \to \infty} \lim_{m \to \infty} g(m,n)$ and $\lim_{m \to \infty} \lim_{n \to \infty} g(m,n)$ are not the same. The former limit represents taking Riemann sums of $f$ on all of $[0,1]$ with the evaluation points pushed closer and closer to the left endpoint. It blows up. The latter limit represents taking the Riemann integral of $f$ on $[1/m,1]$ and then sending $m \to \infty$. It converges.