function which is Riemann integrable

Question

Consider $f:[-1,1]\to\mathbb{R}$, $x\mapsto \begin{cases} 1, & \text{if } x=0 \\ 0 & \text{else } \end{cases}$ I want to know why f is Riemann integrable and I tried something. First of all we had the definition in lecture: $\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}$ and $\int_{-1^*}^{1} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, \phi\le f \}$. Now the bounded function f is Riemann integrable, if $\int_{-1}^{1^*} f(x)\,dx=\int_{-1^*}^{1} f(x)\,dx$.

My first try: consider the step function $\phi:[-1,1]\setminus\{0\}\to\mathbb{R},\; x\mapsto 0$. It is $f=\phi$ everywhere on $[-1,1]\setminus \{0\}$. Therefore it is
$\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}=\int_{-1}^{1} \phi(x)\,dx=0$ and $\int_{-1^*}^{1} f(x)\,dx=\int_{-1}^{1} \phi(x)\,dx=0$. Therefore f is Riemann-integrable and $\int_{-1}^{1} f(x)\,dx=0$. Could you help me, to correct my solution?

I'm not sure if it is ok, because my step function isn't defined in $x=0$. But we say that $\phi:[-1,1]\to\mathbb{R}$ is a step function if you have a partition of the interval $[-1,1]$, $x_0=-1<x_1<..<x_n=1$, such that $\phi_{|(x_{i-1},x_i)}=c_i\in\mathbb{R}$ for $i=1,..,n$.
Maybe I have to define $\phi$ in $x=0$ too but take $x=0$ as one of the $x_i's$.

I will try next to do this with Riemann Sums. Regards

Answer 1

Fix $1> \epsilon > 0$, and consider the step function $$ \varphi(x) = \begin{cases} 1 &: -\epsilon \leq x \leq \epsilon \\ 0 &: \text{ otherwise} \end{cases} $$ Then $$ \int_{-1}^1 \varphi(x)dx = 2\epsilon $$ Hence $$ \int_{-1}^{1^{\ast}} f(x)dx = 0 $$ Now for any step function $\psi \leq f$, then choose a partition $-1 < x_1 < x_2 < \ldots < x_n = 1$ and constants $c_i$ such that $\psi = c_i$ on $(x_{i-1},x_i)$. Now consider the two cases

• $0 = x_i$ for some $i$ or
• $0 \in (x_{i-1},x_i)$ for some $i$

In either case, prove that $$ \int_{-1}^1 \psi(x)dx \leq 0 $$ Since the constant function $0 \leq f$ is a step function, it follows that $$ \int_{-1^{\ast}}^{1} f(x)dx = 0 $$ and so $f$ is Riemann integrable.

Answer 2

A piecewise constant function lies below $f$ if and only if it is nonpositive everywhere. So the supremum of the lower sums of $f$ is zero. Among piecewise constant functions lying above $f$, for any $\delta>0$ we have a function which is $1$ on $[-\delta,\delta]$ and zero elsewhere. The integral of this function is $2 \delta$, which can be made arbitrarily small, so the infimum of the upper sums of $f$ is also zero. We have checked the Darboux criterion for integrability.

What might be left to prove is that the Darboux definition is equivalent to your original definition of Riemann integrability.