Let $f(x)=\small{\begin{cases} 1 \; &\text{if} \, 0\leq x<1 \\ 3 \; &\text{if} \, 1< x\leq 2 \\ \end{cases}}$. Show its integrability.

Question

$$ \text{Let }f(x)=\begin{cases} 1 \quad &\text{if} \, 0\leq x<1 \\ 3 \quad &\text{if} \, 1< x\leq 2 \\ \end{cases}$$

$$\text{Use the Integrability Reformulation to prove that }\int^{2}_{0}f(x)dx \text { is integrable}$$

The integrability reformulation statement says that:

$$\forall \epsilon > 0, \exists \text {partition p of [0,2] s.t }U(f,p)-L(f,p) < \epsilon$$

Usually when I do these questions it is just testing whether $x \in \mathbb{Q}$ or not in it. I never did I question with a interval as such. How do I do something like this?

The only thing I can think of is breaking up the integral, and prove that both parts are separately integral, so their sum must be too. Is this valid?

$$\text{Prove } \int^{1}_{0}f(x)dx + \int^{2}_{1}f(x)dx$$

Answer 1

I would say no. Not because breaking it up is wrong mathematically, but because that is not how you are asked to do it. The object of such exercises is to strengthen your understanding of the indicated method. So making use of other results fails this purpose. After all, how do you know it is okay to split it up into sub-intervals? It is because somebody used the definition of integrability to prove it works. Do you understand that proof? If so, then you can prove this the same way. If you don't have a solid grasp of that proof, then this exercise can give you some practical experience that will improve your understanding.

I am a little flummoxed by your indication that you could handle step functions where the test is $x \in \Bbb Q$, which are wrought with potential issues (indeed, the integrability of such functions differs depending on which theory of integration you are using), but are confused by the much simpler test of $x$ being in certain intervals.

There is one issue with this problem: As you gave it, $f(1)$ is not defined. Of course we know that integration doesn't depend on the value of the function at any one point. However, the definition of Riemann integration usually requires it to be defined at every point. I say "usually" because it can differ from book to book, with some authors preferring a different formulation than others. "$U(f,p), L(f,p)$" is not enough for me to determine what formulation you are using.

So arbitrarily assign some value to $f(1)$, and let $M = |f(1)| + 3$. Now consider the partition $$p = \left\{0, 1-\frac \epsilon{2M},1+\frac \epsilon{2M},2\right\}$$

On the interval $p_0 = [0, 1- \epsilon /2M]$, $\max f = \min f = 1$.

On the interval $p_1 = [1- \epsilon /2M, 1 + \epsilon /2M], \max f = \max\{3, f(1)\}$, while $\min f = \min\{1, f(1)\}$

On the interval $p_2 = [1 + \epsilon /2M, 2]$, $\max f = \min f = 3$.

So $$U(f,p) = 1\cdot(1- \epsilon /2M) + \max\{3, f(1)\}\cdot\frac \epsilon M + 3\cdot(1- \epsilon /2M)\\L(f,p) = 1\cdot(1- \epsilon /2M) + \min\{1, f(1)\}\cdot\frac \epsilon M + 3\cdot(1- \epsilon /2M)$$ And $$U(f,p) - L(f,p) = (\max\{3, f(1)\} - \min\{1, f(1)\})\frac \epsilon M \le M \frac \epsilon M = \epsilon$$