I know that not every bounded function $f$ on an interval $[a,b]$ is Riemann integrable. However, I seem to have shown that every such function is Riemann integrable. I know something must be wrong with my proof, but I can't figure out what it is. Here is what I did.
Let $M$ be a bound of $f$ on $[a,b]$. Thus $|f(x)| \leq M$ for all $x\in[a,b]$. Let $P$ and $Q$ be two arbitrary tagged partitions of $[a,b]$, and let $R(f,P)$ and $R(f,Q)$ be the corresponding Riemann sums, respectively. We have that
$\begin{align*}
|R(f,P) – R(f,Q) | &= \left\vert\sum_{k=1}^{n}f(x_k^*)(x_k-x_{k-1}) – \sum_{k=1}^{n}f(y_k^*)(y_k-y_{k-1}) \right\vert\\
&\leq \sum_{k=1}^{n}|f(x_k^*)|(x_k-x_{k-1}) + \sum_{k=1}^{n}|f(y_k^*)|(y_k-y_{k-1})\\
&\leq \sum_{k=1}^{n}M(x_k-x_{k-1}) + \sum_{k=1}^{n}M(y_k-y_{k-1}).
\end{align*}$
Here, $x_k^*$ and $y_k^*$ are the tags of $P$ and $Q$, respectively. Now, for any $\epsilon > 0$ let $\delta = \frac{\epsilon}{2Mn}$. Then if the norms of $P$ and $Q$ are each less than $\delta$,
\begin{align*}
\sum_{k=1}^{n}M(x_k-x_{k-1}) + \sum_{k=1}^{n}M(y_k-y_{k-1}) &\leq \sum_{k=1}^{n}M\frac{\epsilon}{2Mn} + \sum_{k=1}^{n}M\frac{\epsilon}{2Mn} \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\end{align*}
This implies that the partitions satisfy the Cauchy criterion and thus $f$ is Riemann integrable.
Where am I going wrong?
Best Answer
You made the assumption that all tagged partitions have the same number of points $n$. But if you want the norm of your partition to be arbitrarily small, then $n$ will grow larger.