Is f Riemann integrable on $[0,2] $ ? Yes/No

Question

Is the function $f(x) = \begin{cases} \text{1 if} \ x \neq 1 \\ 0\ \text {if x = 1} \end{cases}$ Riemann integrable on $[0,2] $ ?

My attempt : I think No , because here $L(P,f) \neq U(P,f)$

Is it True?

Answer 1

$f$ is bounded on $[0,2]$ and continuous on $[0,2] \setminus \{1\}.$ Hence, $f$ is Riemann- integrable.

Answer 2

The function is Riemann integrable. Even though the maximum and minimum in one/two of the intervals in any partition are different,this difference vanishes when you let the lengths of the intervals in the partition tend to $0$.

Answer 3

We can show the following: Let $a,b\in\mathbb R,a<b$ and $f:[a,b]\rightarrow \mathbb R$ Riemann-integrable. Let $g:[a,b]\rightarrow \mathbb R$ such that $M=\{x\in[a,b];f(x)\neq g(x)\}$ is finite. Then $g$ is Riemann-integrable with $\int\limits_{a}^{b}\!f(x)\,\mathrm{d}x=\int\limits_{a}^{b}\!g(x)\,\mathrm{d}x$. (Notice that we do not need $f$ or $g$ to be continuous).

Proof: Let $\varepsilon>0$ und $I=\int\limits_{a}^{b}\!f(x)\,\mathrm{d}x$, then there exist step-functions $\varphi,\psi:[a,b]\rightarrow\mathbb R$ such that $$\varphi\leq f\leq\psi,~I-\int\limits_{a}^{b}\!\varphi(x)\,\mathrm{d}x\leq\varepsilon,~\int\limits_{a}^{b}\!\psi(x)\,\mathrm{d}x-I\leq\varepsilon$$ We define $$\varphi^*(x)=\begin{cases} g(x), & x\in M \\ \varphi(x), & x\notin M\end{cases},\quad \psi^*(x)=\begin{cases} g(x),&x\in M \\ \psi(x),&x\notin M\end{cases}.$$ Because $M$ is finite these are again step-functions and we can easily see that $$\varphi^*\leq g \leq \psi^*,~ \int\limits_{a}^{b}\!\varphi(x)\,\mathrm{d}x=\int\limits_{a}^{b}\!\varphi^*(x)\,\mathrm{d}x,~\int\limits_{a}^{b}\!\psi(x)\,\mathrm{d}x=\int\limits_{a}^{b}\!\psi^*(x)\,\mathrm{d}x.$$ Therefore we have $$I-\int\limits_{a}^{b}\varphi^*(x)\,\mathrm{d}x\leq \varepsilon,\quad \int\limits_{a}^{b}\psi^*(x)\,\mathrm{d}x-I\leq\varepsilon$$ from which we conclude that $g$ is Riemann-integrable with $\int\limits_{a}^{b}\!g(x)\,\mathrm{d}x=I$.