Proving a function has the following integral

Question

the function is defined as follows:

$g : [1, 3] \rightarrow\mathbb{R}$ s.t $~~g(x) = 0$ if $x \neq 2$ and $= 1$ if $x = 2$

I have to prove that the Riemann integral of the above function from $1$ to $3$ is $ = 0$.

I'm trying to prove that it's Riemann Steiltjis integrable and taking the function $α = x$. What I have done so far is this:

let $α(x) = x $ increasing and continuous at $x = 2$.

then $\forall \epsilon \gt 0, \exists \delta \gt 0 ~~s.t.~~ |x-2| \lt \delta ~~implies~~ |\alpha(x) - \alpha(2)| = |x-2| < \epsilon $

let p be a partition of $[1, 3]$.

then $\exists i_0~~s.t. ~~ 2 \in [ x_{i_0 -1}, x_{i_0}]$

But then what? what can i say about $U(p, g, \alpha) - L(p, g, \alpha)$ ?

I'd appreciate any help.

Answer 1

To show that the integral equals zero, you can do the following.

I think you are familiar with $\overline{\int g(x)dx}:=\inf\limits_{P}{U(P,g)}$ and $\underline{\int g(x)dx}:=\sup\limits_{P}{L(P,g)}$

It is enough to prove that $\overline{\int g(x)dx}=\underline{\int g(x)dx}=0$, because if $g(x)$ is riemann integrable, what can you say then about $\int g(x)dx$?

Let's prove that: given $\epsilon>0$, we want to find a partition $P$ such that: \begin{align} U(P,g)-\overline{\int g(x)dx}<\epsilon \hspace{10pt} \text{and} \hspace{10pt} L(P,g)-\underline{\int g(x)dx}>-\epsilon \hspace{15pt} \text{(why?)} \end{align}

I only show that $\overline{\int g(x)dx}=0$, I let you do the other one.

Choose partition $P$, such that $|P|<\epsilon$, $|P|:=\max{\{\Delta x_k | k \in [1, n]\}}$.

$U(P,g)=\sum\limits_{k=1}^n \sup\limits_{x\in[x_{k-1},x_{k}]}{g(x)}\Delta x_k=0+ .... + 0 + 1*\Delta x_j +0 +... +0\leq 1*|P|<\epsilon$

( $j $ is between 1 and $n$)

I hope this helps you!

Answer 2

Pick any partition $\mathcal{P} = \{ x_0, \dots, x_n \}$ of $[1,3]$. The Riemann sums over this partition for the function $g$ is:

$$ S_{-}(g,\mathcal{P}) = \sum_{i=0}^{n-1} \inf_{[x_i, x_{i+1}]}g(x) \cdot (g(x_{i+1})-g(x_i))$$

$$S_{+}(g, \mathcal{P}) = \sum_{i=0}^{n-1} \sup_{[x_i, x_{i+1}]}g(x) \cdot (g(x_{i+1})-g(x_i))$$

Obviously $S_{-}(g, \mathcal{P}) = 0$ for all $\mathcal{P}$. Though $S_{+}(g, \mathcal{P})$ could be non zero for some partition $\mathcal{P}$ we must take the $\inf$ over $\mathcal{P}$ of $S_{+}(g,\mathcal{P})$ and this it zero because we can take $\mathcal{P} = \{1,3\}$. Then

$$\sup_{\mathcal{P}}S_{-}(g,\mathcal{P}) = \inf_{\mathcal{P}}S_{+}(g,\mathcal{P}) = 0$$