calculus - How to prove $f(x)$ is integrable function on $[0,2]$

Let

$$ f(x) = \begin{cases}
1, & 0 \le x < 1 \\
0, & x=1 \\
1, & 1
\end{cases}$$

How Prove that $f(x)$ integrable on $[0,2]$

I know that $f(x)$ defined and Bounded on [a,b] then $f(x)$ integrable

$iff$

$\forall ϵ>0$ there is Partition $P$ so that

$$U(f;P)-L(f;P)<ϵ$$

how can i prove that?

thanks

Answer

Notice that on any interval $(a,b)\subset [0,2]$ we have that $\max_{x\in (a,b)}{f(x)}=1$ and $$\min_{x\in (a,b)}f(x)=\begin{cases}
0 & \mbox{ if } 1\in (a,b),\\
1 & \mbox{ otherwise.}
\end{cases}$$

Let $n\in \mathbb{N}$ and let $P=(P_i)_{i\in I}$ be any partition such that $P_j=(1-\frac{1}{n},1+\frac{1}{n})$ for some $j$. Clearly $U(f;P)=2$. Notice that $L(f;P)=2-\frac{2}{n}$. Since $n$ was chosen arbitrarily, we can get $U(f;P)-L(f;P)$ as small as we want.