# Let $f:[a,b]\to\Bbb{R}$ be a bounded function which is continuous except at a point $x_0\in [a,b]$. Then, $f$ is Riemann integrable over $[a,b]$.

Let $$f:[a,b]\to\Bbb{R}$$ be a bounded function which is continuous except at a point $$x_0\in [a,b]$$. Then, $$f$$ is Riemann integrable over $$[a,b]$$.

My Proof

Consider the restriction

\begin{align} g:[a&,b]\backslash\{x_0\}\to\Bbb{R} \\ &x\mapsto g(x)=f(x) \end{align} Since $$f$$ is bounded, then $$g$$ is also bounded on $$[a,b]$$. So, there exists $$K\geq 0,$$ such that $$|g(x)|\leq |f(x)|\leq K,\;\;[a,b].$$ Since $$\;g:[a,b]\backslash\{x_0\}\to\Bbb{R}$$ is continuous and bounded where $$x_0\in [a,b]$$, then $$g$$ is Riemann integrable over $$[a,b]\backslash\{x_0\}$$.

As a result, there exists partition $$P$$ of $$[a,b]$$ such that \begin{align} \| P \| <\dfrac{\epsilon}{8K},\;\;U(g,P)<\int^{b}_{a}g(x)dx+\dfrac{\epsilon}{2}\;\;\text{and}\;\;L(g,P)>\int^{b}_{a}g(x)dx-\dfrac{\epsilon}{2}.\end{align} Let $$P_0=\{x_0\}\cup P:=\{t_0,t_1,\cdots ,t_n\},\;\;A=\{j:[t_{j-1},t_{j}]\cap \{x_0\}\neq \phi\}\;\;\text{and}\;\;$$ \begin{align} B=\{j:[t_{j-1},t_{j}]\cap \{x_0\}= \phi\}.\end{align} The set $$\{x_0\}$$ can only be in at most two subintervals, and so, \begin{align}U(f,P_0)&=\sum_{j\in A}M_j(f)\Delta x_j+\sum_{j\in B}M_j(f)\Delta x_j\\&=\sum_{j\in A}[M_j(f)-M_j(g)]\Delta x_j+U(g,P_0)\\&\leq 2K\sum_{j\in A}\Delta x_j+U(g,P_0)\\&\leq 2K\| P \|+U(g,P) \\&<\int^{b}_{a}g(x)dx+\epsilon\end{align} Lower Darboux sums follow in a similar way. Hence, we have that

\begin{align} \int^{b}_{a}f(x)dx=\int^{b}_{a}g(x)dx.\end{align} Question: Is this proof correct? If no, what did I do wrong? You can also provide me a better proof.

• It's hard to say without the text from which you are working, but my guess is that you are glossing over the integrability of $g$ too quickly. Since $g$ is not defined on $[a, b]$ but rather on $[a, x_0) \cup (x_0, b]$, depending on your definitions it might not even make sense to say $g$ is integrable on that set, its integrability might not follow from its continuity and boundedness, and finally its integrability might not imply the existence of your $P$. However, the idea of your solution definitely seems to be in the right direction. – Mees de Vries Jan 24 at 13:54
• @Mees de Vries: Thanks for that! Is it possible to see another kind of solution? – user560896 Jan 24 at 14:27
• Hint: choose some small $\eta > 0$, based on $\epsilon$ and $K$, and then consider the two functions $g_1, g_2$ which are the function $f$ restricted to $[a, x_0 - \eta], [x_0 + \eta, b]$, respectively. – Mees de Vries Jan 24 at 14:31

You don't need to introduce $$g$$. Without any loss of generality one can assume $$x_0$$ to be an end point of $$[a, b]$$ say $$x_0=a$$. Consider $$c=a+(\epsilon/4K)$$ and then $$f$$ is uniformly continuous on $$[c, b]$$ and hence we have $$\delta>0$$ such that $$|f(x) - f(y) |<\frac{\epsilon} {2(b-a)}$$ whenever $$x, y\in[c, b]$$ with $$|x-y|<\delta$$. Let $$P'$$ be a partition of $$[c, b]$$ with norm $$||P'||<\delta$$ and let $$P=\{a\}\cup P'$$. Then we have $$U(P, f) - L(P, f) <2K\cdot\frac{\epsilon} {4K}+U(P',f)-L(P',f)<\epsilon$$ and we are done.