I need to prove the following:
Suppose $g:[a,b] \to \mathbb{R}$ is bounded and continous except at $x_{1},...,x_{n} \in [a,b]$. Prove that $g$ is Riemann integrable on $[a,b]$.
I know that this question is duplicated, but the only thing I want to know if this is right. Thanks for your help.
My attempt:
$g$ is bounded $\Rightarrow \exists M \in R$ such that $g<|M|$. For all $\varepsilon>0$ :
Is $D=\{d_0,...,d_k\}$ the set of discontinuous points of $g$ and $P=\{x_0,...,x_n\}$ partition of $[a,b]$ containing $D$ such that:
- the sum of intervals determined by $P$ such that one of the points is of discontinuous of $g$ is $\sum_{D}^{}(x_{i}-x_{j})<\frac{\varepsilon}{4M}$
- $K=\{k_0,...,k_j\}$ is the set of other points of $P$ such that $\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})<\frac{\varepsilon}{2}$ (in this points $f$ is continuous and we can make this choice).
So: $$ U(g, K \cup D)-L(g,K \cup D)=\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+\sum_{D}^{} (M_i-m_i) (x_{i}-x_{j}) $$ $$ <\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+2M \sum_{i=1}^j (x_{i}-x_{i-1})<\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+2M \frac{\varepsilon}{4M} $$
$$ =\sum_{i=1}^n (M_i-m_i) (k_{i}-k_{i-1})+\frac{\varepsilon}{2} <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} < \varepsilon $$
From that: $g$ is Riemann Integrable.
I used the answer of @Felipe, in How do I prove that a function with a finite number of discontinuities is Riemann integrable over some interval? I think is right but I don't know if I can affirm if the bold part, Can you help me with the verification of this please. thank you a lot. :)
