prove that $f$ is Riemann integrable on $[0,1]$

Question

Considering the function $f$ defined on $[0,1]$ by

$$ { f }(x)=\begin{cases} 0\quad ,\quad x \in [0,1) \\ 1\quad ,\quad x=1\end{cases} $$

I need to prove that $f$ is Riemann integrable on $[0,1]$

So,

Let $\epsilon>0.$ Let $P_{\epsilon}$ be a partition of the interval $[0,1]$ where $P_{\epsilon}=\{[0,1-2\epsilon], [1-2\epsilon,1-\epsilon],[1-\epsilon,1]\}$ for some fixed $0<\epsilon<1$

And I know that,

$$U(f,P) = \sum_{[x_i,x_{i+1}] \in P} (x_{i+1} - x_i) \sup_{[x_{i+1},x_i]} f(x)$$

$$ L(f,P) = \sum_{[x_i,x_{i+1}] \in P} (x_{i+1} - x_i) \inf_{[x_{i+1},x_i]} f(x) $$

Then how can I show that $L(f,P)=U(f,P)$ which would mean it would be Riemann integrable.

Answer 1

It is a fairly easy result that the Riemann Integral exists if $U(f,P) - L(f,P) = \epsilon$ for some Partition, $P$

Also, note that My upper sum here must be $\epsilon$:

$\begin{equation*} \begin{split} \sum_{[x_{i},x_{i+1}]} (x_{i + 1} - x_{i}) \sup_{[x_{i},x_{i+1}]} f(x) &= \big((1 - 2\epsilon) - 0)\cdot 0 + \big((1 - \epsilon) - (1 - 2\epsilon)\big)\cdot 0 + \big((1-(1 - \epsilon)\big)\cdot 1 \\ &= 0 + 0 + (\epsilon)\cdot 1 \\ &= \epsilon \end{split} \end{equation*}$ because we are taking the supremum of $f(x)$ in each subinterval, but $f(x)$ is uniformly zero for all points in my first two subintervals.

Pretty clearly my lower sum must be $0$, and so $U(f,P) - L(f,P) = \epsilon - 0 = \epsilon$

Answer 2

Consider $f_n=0$ and $g_n$ whose restriction on $[0,1-1/n]$ is zero and whose restriction on $[1-1/n,1]$ is $1+1/n$, $f_n\leq f\leq g_n$ and $\mu(g_n-f_n)\leq {1\over{n}}$.