Continuity of a Function $f$

Question

I've been studying different types of functions and I came across one on What is an example that a function is differentiable but derivative is not Riemann integrable, but I can't figure out why $f(x)=x^{ \frac{3}{2} }sin(\frac{1}{x})$ on $[0,1]$ is continuous, because it seems that that it doesn't exist at $x=0.$ But I know it is differentiable on $(0,1),$ and not Riemann integrable. Some clarification, please?

Answer 1

At first lets take a look at the theoretical asspects of the subject and let me begin with some basic definitions in order to finally define in a rigorous way the concept of Riemann Integral.

• Partitions

A partition $\mathcal{P}$ of an interval $[a, b]$ is a finite sequence of real numbers of the form $$a = x_0 < x_1 < x_2 < \cdots < x_n = b$$ Each $[x_i, x_{i+1}]$ is called a subinterval of the partition.

Norm of a partition is defined to be the length of the longest subinterval, that is, $$ \left\lVert \mathcal{P} \right\rVert = \max_{i=1,\cdots n} (x_{i+1}-x_i), \quad i \in [0,n-1]$$

Finally, the partition $\mathcal{P^*}$ of the interval $[a, b]$ it is a refinement of $\mathcal{P}$ if $$\mathcal{P} \subset \mathcal{P^*} $$

Let's see a example of a refinement.

Consider the interval $[0,1]$, one partition of the interval is $$0<\frac{1}{3}<\frac{1}{2}<1$$ another one is

$$ \color{red}{0}<\frac{1}{9}<\frac{1}{4}<\color{red}{\frac{1}{3}}<\color{red}{\frac{1}{2}}<\frac{2}{3}<\frac{7}{9}<\color{red}{1}$$

The second one is a refinement of the first.

A collection of points associated with the partition $\mathcal{P}$, is a finite sequence of real numbers $\mathit{T}_{\mathcal{P}}$ or more simple $\mathit{T}=\{t_1, \cdots t_{n-1}\}$ such that $$t_i \in [x_i, x_{i+1}]$$

with $[x_i, x_{i+1}]$ the subinterval of the partition.

now we could define the Riemann sums.

• Riemann sums

Let $f:[a,b]\to \mathbb{R}$ be a function defined over a closed and bounded interval.

The Riemann sum of $f$ associated with the partition $\mathcal{P}$ and the collection of points $\mathit{T}$ it is defined to be $$\mathrm{S}(f,\mathcal{P},\mathit{T})= \sum_{i=1}^{n-1}f(t_i)(x_{i+1}-x_i)$$

Notice two things:

1. The Riemann sums approximate the area beneath the graph of $f$ (when $f$ is everywhere non negative)

2. and moreover this approximation becomes better and better as the norm of the partition tents to $0$

and finally...

• Riemann Integrability

Definition Let $f:[a,b]\to \mathbb{R}$ be a function over a closed and bounded interval.

$f$ is Riemann integrable if $$\lim_{\left\lVert \mathcal{P} \right\rVert \to 0}\mathrm{S}(f,\mathcal{P},\mathit{T})< \infty, \quad \forall \text{ collection } \mathit{T}$$

So if the limit exists we define the Riemann integral of $f$ to be $$\int\limits _a^b f:=\lim_{\left\lVert \mathcal{P} \right\rVert \to 0}\mathrm{S}(f,\mathcal{P},\mathit{T})$$

In order to have a complete view of the topic let me say that, Riemann sums is one way two define rigorously the Riemann integral another way is (Upper and Lower)Darboux sums which are, in my opinion, is for most purposes technically easier to work with and thus is the one which is carefully developed in most undergraduate texts. But it has been proven that the two definitions are equivalent.

At last let my note that

$f$ is Riemann integrable $\implies$ $f$ is bounded

This is something that could be concluded easily by insight, but here you could find a more formal discussion

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So now it is clear that before we examine that a function is Riemann integrable first we have to ensure that the function

• is defined over a closed and bounded interval
• and is bounded

And as result a priori by the definition the function $f: (0,1] \to \mathbb{R}$, with $$f(x)=x^{ \frac{3}{2} }\sin(\frac{1}{x})$$ it is not Riemann integrable.

Those two prerequisites, for the function under integrability examination, come up really naturally if we consider the motivation of the study of the integral, which is to calculate the area bellow a graph of a function.

How to calculate the area between a graph and a infinite interval? Or how could a unbounded region have finite area? But in order to answer those questions we defined a more general integral the improper Reimann integral.

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Forget for a while the $f$ and consider the function $g:[0,1] \to \mathbb{R}$, with the below formula $$g(x)=\begin{cases} f(x), &\quad \text{ if } x \in (0,1] \\ 0, &\quad \text{ if } x= 0\end{cases}$$

You can check easily that $g$ is continuous, so that implies that $g$ is Riemann integrable (since $g$ is a continuous function defined over a compact interval)

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Although $f$ doesn't feet the criteria (it is not defined over a closed and bounded interval), $f$ is bounded and moreover our insight (if we take a look at its graph) it tells us that it should be integrable.

Remember that the Riemann integral (when it exists) is a good way to evaluate the area below a graph,

since $g$ is integrable over $[0,1]$ the area between the graph of $g$ and the $x$-axes, from $x_0=0$ and $x_1=1$, is

$$\int\limits _0^1 g$$

Now lets use our insight and try to imagine what is the area between the graph of $f$

We have that $\displaystyle\lim_{x \to 0} f(x)=0$ this means that $f$ is bounded near $x_0=0$ (it is not explodes to $\pm \infty$)

Moreover $f$ is defined over $(0,1]$ and $\forall x \in (0,1], \, \, f(x) =g(x)$

So it is logical to conclude that the area we are looking for equals $\displaystyle\int\limits _0^1 g$

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To make things clear hear is a major result of Real Analysis, the Theorem below characterizes the class of Riemann integrable functions.

Theorem (Lebesgue Criterion)

A function f on a compact interval $[a, b]$ is Riemann integrable $\iff$

1. it is bounded

2. it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

Last but not least let me state the following.

A countable collection of points in $\mathbb{R}$ has Lebesgue-measure zero (or in simple worlds does not have any length)