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We have real piecewise function $f(x)$ such that: $$f(x)=\begin{cases} x&\text{if $x\geq 0$,}\\ 1-x&\text{if $x<0$.} \end{cases}$$ We define $L_n$ as the lower Riemann sum for a given interval by breaking the interval to $n$ partitions. For example for $n =2$ and $[0,2]$ the partitions are $[0,1],[1,2]$.

We want to find a general equation for $L_{2n}$ of $f(x)$ in interval $[-1,1]$ in terms of $n$. In the book I saw this problem the answer was $L_{2n} = 2-\frac{1}{n}$. in the solution it said we have $n$ intervals between $[-1,0]$ and $n$ intervals between $[0,1]$. then it calculated the sums and got the above equation. My problem is that in the solution, for $x=0$ in the left interval, it said $\inf f(x) = 1$ and for the right interval it said $\inf f(x) =0$. I think it is wrong. In the definition of lower sum, it is said that we must use infimum of $f(x)$ in the (closed) interval $[a,b]$ so in the interval $[-\frac{1}{n},0]$ I think we must use $0$ for $\inf f(x)$.

If we use the method given in the book, for $n =1$ or $2n=2$, the lower sum would be $L_{2*1} = 2-\frac{1}{1} = 1$ but if we use the method I said, it would be $0$. because $f(-1) = 2, f(0) = 0 , f(1) = 1$ so the infimum in both intervals are $0$ and $\frac{2}{2}(0+0) = 0$

So, which method is right?

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The evaluation of this Riemann lower sum $L_{2n}$ depends on its definition.

Let $x_j=j/n$ with $j=-n,\dots, n$ is a uniform partition of the interval $[-1,1]$ and note that $f$ is decreasing in $[-1,0]$ and increasing in $[0,1]$. Let $f_1(x)=1-x$ and $f_2(x)=x$.

1) The infimum is taken over the half-closed interval $[x_{j-1},x_j)$: $$ \begin{align} L_{2n}&=\frac{1}{n}\sum_{j=-n+1}^{n}\inf_{t\in [x_{j-1},x_j)}f(t) \\ &=\frac{1}{n}\sum_{j=-n+1}^{0}f_1(x_j)+\frac{1}{n}\sum_{j=0}^{n-1}f_2(x_j)=1-\frac{1}{n^2}\sum_{j=-n+1}^{0}j+\frac{1}{n^2}\sum_{j=0}^{n-1}j\\ &=1+\frac{1}{n^2}\sum_{j=0}^{n-1}j+\frac{1}{n^2}\sum_{j=0}^{n-1}j=1+\frac{n(n-1)}{n^2}=2-\frac{1}{n}. \end{align}$$

2) The infimum is taken over the closed interval $[x_{j-1},x_j]$: $$\begin{align} L_{2n}&=\frac{1}{n}\sum_{j=-n+1}^{n}\inf_{t\in [x_{j-1},x_j]}f(t) \\ &=\frac{1}{n}\sum_{j=-n+1}^{-1}f_1(x_j)+\frac{f_2(x_0)}{n}+\frac{1}{n}\sum_{j=0}^{n-1}f_2(x_j)=\frac{n-1}{n}-\frac{1}{n^2}\sum_{j=-n+1}^{-1}j+\frac{1}{n^2}\sum_{j=0}^{n-1}j\\ &=\frac{n-1}{n}+\frac{1}{n^2}\sum_{j=1}^{n-1}j+\frac{1}{n^2}\sum_{j=0}^{n-1}j=\frac{n-1}{n}+\frac{n(n-1)}{n^2}=2-\frac{2}{n}. \end{align}$$

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  • $\begingroup$ I know that. But I think we must not use x=0 in the decreasing function. because we know that the infimum of that in $[-1/n , 0]$ is 0. not 1. $\endgroup$ – titansarus Mar 17 '18 at 8:45
  • $\begingroup$ @titansarus Is it clear now? $\endgroup$ – Robert Z Mar 17 '18 at 8:57
  • $\begingroup$ It is clear but it still uses $f_1(x_0)$. For example, evaluate this for n =1, if we we use the above we get 1, but if we use $f(-1) =2, f(0)=0,f(1) = 1$ we get 0+0=0. I think if use exactly the definition, we don't get $2-1/n$. $\endgroup$ – titansarus Mar 17 '18 at 9:01
  • $\begingroup$ I see your point. It depends how the lower sum is defined. In particular if the interval where we take the infimum is open or closed. What is the definition in your textbook? $\endgroup$ – Robert Z Mar 17 '18 at 9:01
  • $\begingroup$ It is exactly the definition in wikipedia for lower riemann sum, en.wikipedia.org/wiki/Riemann_sum $\endgroup$ – titansarus Mar 17 '18 at 9:05

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