Is this riemann integrable?

Question

So as far as I know to be riemann integrable there should be an value (lets say e>0 ) that $U(f,P)-L(f,P) < e$. However my teacher showed that below function is not riemann integrable by this method.

Method

$$f(x) = \begin{cases} 1 & x\in[0,1]\cap \Bbb Q , \\ 0 & otherwise .\end{cases}$$

What she did was, $U(f,P) = 1(1-0)$ and $L(f,P) = 0(1-0)$ Since both are not equal this is not riemann integrable.

Problem

What she did was clear to me. So I tried to prove this beliw one using the same logic. But I don't know how to start it. If someone can please tell me what should I do to prove whether this is reimann integrable or not.

Let,

$$f(x) = \begin{cases} 1 & x\in[0,1]\setminus\ \frac{1}{2}\, \\ 0 & x=\frac{1}{2} .\end{cases}$$

Answer 1

Let $P$ be a partition and let $a$ be the length of the sub-intervals which includes $1/2$.

Note that: if $1/2$ is an end point, then there exist two such sub-intervals and if it is not, then there exists one sub-interval.

Since each sub-interval has image value $1$, we have the supremum on each interval is $1$. Therefore $$U(f,P)=1.$$ Since the length of interval having the $1/2$ is $a$, we have $$L(f,P)=1-a.$$

Now $$U(f)=1$$ and $$L(f)=\sup\{ 1-a \; | \; a \in (0,1]\}=1.$$ Therefore, it is Riemann integrable and the integral value is $1$.

Answer 2

It's a good idea to actually draw a few Riemann approximations, in this case.

Suppose $n$ is "large", and let's cut $[0, 1]$ into $n$ equal subintervals. Then:

• The upper Riemann sum will have every single rectangle of height $1$ (why?), so will guess that the area is $(1\cdot {1\over n})\cdot n=1$.

• The lower Riemann sum will have all but one rectangle of height $1$, and one will be of height $0$ (why?). So the upper Riemann sum will be $(1\cdot{1\over n})\cdot (n-1)+ 0\cdot{1\over n}={n-1\over n}$.

Now, do these Riemann sums look like they're converging to the same value as $n$ goes to infinity? What does that suggest about whether $f$ is Riemann integrable?

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The distinction between your $f$ and the function your teacher showed you (the Dirichlet function) is that your $f$ only has one "bad" point ($f$ is discontinuous at $x={1\over 2}$), whereas the Dirichlet function has lots (every point is a bad point).