I am trying to follow the proof at the page http://en.wikipedia.org/wiki/Riemann_integral which shows how a Darboux Integrable function is Riemann Integrable also. In general a partition of an interval $[a,b]$ is picked as $y_0,y_1,...,y_m$ such that its upper and lower Darboux sums are in $\epsilon/2$ distance from the integral $s$. Also the following are set:

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(assume $r>0$) and as the mesh (longest subinterval) of a new partition $x_0,x_1,...,x_n$ of the interval $[a,b]$ a $\delta$ is picked as:

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The purpose is to show that the Riemann sum of the interval $x_0,x_1,...,x_n$ is within $\epsilon$ distance of $s$. The proof for first case when each subinterval $[x_i,x_{i+1}]$ of the second partition is contained in a subinterval $[y_j,y_{j+1}]$ of the first partition is very clear. The problematic one is with the case when a subinterval $[x_i,x_{i+1}]$ spans across the boundary of a subinterval $[y_j,y_{j+1}]$ such that $y_j < x_i < y_{j+1} < x_{i+1} < y_{j+2}$. Here, the term $f(t_i)(x_{i+1}-x_i)$ in the Riemann sum is splitted such that: $f(t_i)(x_{i+1}-x_i) = f(t_i)(x_{i+1}-y_{j+1}) + f(t_i)(y_{j+1}-x_i)$. There may be at most $m-1$ such splits. If $t_i \in [x_i,y_{j+1}]$, then the second term $f(t_i)(y_{j+1}-x_i)$ is already bounded by the Darboux sum such that $M_j > f(t_i) > m_j$. Then the other terms should be shown as bounded.

My own approach here was that: As the worst case assume that there are $m-1$ splits. At $m-1$ subintervals, the Darboux sum fails to bound the Riemann sum. Let the upper Darboux rectangle, lower Darboux rectangle and the Riemann rectangle, respectively, at these problematic subintervals as: $M_k \Delta x_k ,m_k \Delta x_k, f(t_k) \Delta x_k$, $(1 \leq k \leq m-1)$. To bound the Riemann sum, I subtract the Darboux rectangles for these subintervals and add the Riemannian ones ($U$ is the upper Darboux sum, $L$ is the lower, R is the Riemann sum):

$$ U - \sum_{k=1}^{m-1} M_k \Delta x_k + \sum_{k=1}^{m-1} f(t_k) \Delta x_k > R > L - \sum_{k=1}^{m-1} m_k \Delta x_k + \sum_{k=1}^{m-1} f(t_k) \Delta x_k$$

Now, according to the Wikipedia proof, by using the facts that $U - s < \epsilon /2$, $s - L < \epsilon /2$ and that we have picked a $\delta$ such that $\Delta x_k < \delta < \dfrac{\epsilon}{2r(m-1)}$ for each $(1 \leq k \leq m-1)$ I should be able to show that the Riemann sum, $R$ is in a distance $\epsilon$ from the integral value $s$. But I keep fail to show that. Either I misunderstood the Wikipedia proof or I made a thought error somewhere but I cannot see that. I need help here. I hope it was not too messy since it is a complex proof, somewhat.

Thanks in advance.

  • $\begingroup$ In the Wikipedia proof, I think their Darboux sums and Darboux rectangles are always with respect to the original $y$-partition. It seems you are looking at Darboux rectangles in the $x$-partition. If so, I don't think you can relate the Darboux sum under the $x$-partition back to $\epsilon$ which bounds the Darboux sum under the $y$-partition. This is why they split the Reimann rectangle for $[x_i,x_{i+1}]$ into two pieces: the one containing $t_i$ can be bounded by the Darboux rectangles in interval $[y_j,y_{j+1}]$ and the other between $0$ and $\frac{\epsilon}{2(m-1)}$. $\endgroup$ – Mick A Sep 9 '14 at 16:23

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