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Let $f:[a,b]\to\Bbb R$ be a Riemann-integrable function. Prove that for each $\sigma\gt0$ there exists a partition $\mathcal P$ of $[a,b]$ into congruent sub-intervals(that is, $x_{j}=a+{j(b-a)\over N}$ for $j=0,1,2,\dots,N$ for some integer $N\ge1$) such that $U(f,\mathcal P)-L(f,\mathcal P)\lt\sigma$.

Actually, I know it is true if the question doesn't have strict condition (congruent sub-intervals), but I am stuck in proving this version.

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  • $\begingroup$ U(f,p) is the upper bound of the integral for the partion P,and L(f,p) is the lower bound of the integral for the partition P $\endgroup$ – user144600 Apr 15 '15 at 1:53
  • $\begingroup$ So you know the less strict version is true, can you use that to construct an $N$ such that the strict version is true? $\endgroup$ – abiessu Apr 15 '15 at 1:58
  • $\begingroup$ I tried it, but I still cannot figure out how? $\endgroup$ – user144600 Apr 15 '15 at 2:00
  • $\begingroup$ Suppose we know that a given partition $\mathcal P$ with non-congruent sub-intervals has an interval of smallest size. Can you prove that there exists an $N$ that makes all intervals congruent and at least this small? $\endgroup$ – abiessu Apr 15 '15 at 2:01
  • $\begingroup$ yes,but we coudn't make sure all the end points of the intervals of the orginal partition are included in the endpoints of the intervals of the new partition,no matter how big N is. So that's the problem I want to solve $\endgroup$ – user144600 Apr 15 '15 at 2:10
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With a Riemann-integrable function $f:[a,b]\to\Bbb R$, we are given that we know for every $\sigma\gt 0$ there exists a partition $\mathcal P$ with $N$ sub-intervals $[x_0,x_1],[x_1,x_2],\dots,[x_{N-1},x_N]$ such that $U(f,\mathcal P)-L(f,\mathcal P)\lt\sigma$. Let such a $\sigma_0,\mathcal P_0,N_0$ be given satisfying these conditions for a given $f$. The Upper and Lower sums arise from the Darboux integral, and are defined as follows:

$$\begin{align}M_i &= \sup_{x\in[x_{i-1},x_i]}f(x)\\ m_i &= \inf_{x\in[x_{i-1},x_i]}f(x)\\ U(f,\mathcal P_0)&=\sum_{i=1}^{N_0}(x_i-x_{i-1})M_i\\ L(f,\mathcal P_0)&=\sum_{i=1}^{N_0}(x_i-x_{i-1})m_i\end{align}$$

Putting these together, we have

$$\sigma_0\gt U(f,\mathcal P_0)-L(f,\mathcal P_0)=\sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)$$

and note we also have $\forall i,M_i\ge m_i$, and therefore $\forall i, (x_i-x_{i-1})(M_i-m_i)\lt \sigma_0$.

Our main concern is that the overall value of $U(f,\mathcal P_i)-L(f,\mathcal P_i)$ be maintained or reduced relative to $U(f,\mathcal P_0)-L(f,\mathcal P_0)$ with any partition $\mathcal P_i$ that we choose. In particular, we know that creating new sub-intervals of each interval in the given partition $\mathcal P_0$ produces a refinement of $\mathcal P_0$ and reduces or preserves the difference between the upper and lower sums.

With this in mind, define a new set of values as follows:

$$\begin{align}\delta&=\inf_{i\in[1,N_0]}(x_i-x_{i-1})\\ \lambda&=\sup_{x\in[a,b]}f(x)-\inf_{x\in[a,b]}f(x)\\ \kappa&=\left[\frac 1{\sigma_0-U(f,\mathcal P_0)+L(f,\mathcal P_0)}\right]+1\\ N_1&=(b-a)^2\frac{\kappa\lambda}{\delta}\\ y_k&=a+\frac{k(b-a)}{N_1}\\ \mathcal P_1&=\{[y_0,y_1],[y_1,y_2],\dots,[y_{N_1-1},y_{N_1}]\}\\ L_k&=\sup_{y\in[y_{k-1},y_k]}f(y)\\ \ell_k&=\inf_{y\in[y_{k-1},y_k]}f(y)\\ U(f,\mathcal P_1)&=\sum_{i=1}^{N_1}(y_i-y_{i-1})L_i\\ L(f,\mathcal P_1)&=\sum_{i=1}^{N_1}(y_i-y_{i-1})\ell_i\end{align}$$

With this as the definition of a partition $\mathcal P_1$, we have guaranteed that every interval of partition $\mathcal P_0$ has at least one cut point either inside it or at each endpoint in our new partition (by factor $\frac{b-a}{\delta}$). The factor $\lambda$ ensures that the maximum possible upper and lower values are accounted for. We cannot guarantee that $U(f,\mathcal P_0)-L(f,\mathcal P_0)\gt 0$, but we can say that $U(f,\mathcal P_1)-L(f,\mathcal P_1)\lt \sigma_0$ by factor $\kappa$. Therefore, we have constructed the required partition with congruent intervals and such that $U(f,\mathcal P_1)-L(f,\mathcal P_1)\lt \sigma_0$. Note that we have $y_{k+1}-y_k=a+\frac{(k+1)(b-a)\delta}{\kappa(b-a)^2\lambda}-\left(a+\frac{k(b-a)\delta}{\kappa(b-a)^2\lambda}\right)=\frac{\delta}{\kappa(b-a)\lambda}$.

So now all that remains is to evaluate $\mathcal P_1$ in light of $\mathcal P_0$ and $\sigma_0$ to demonstrate the correctness of our partition. Recall that $U(f,\mathcal P_0)-L(f,\mathcal P_0)\lt \sigma_0$, i.e.,

$$\sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)\lt\sigma_0$$

and therefore consider

$$\begin{align}U(f,\mathcal P_1)-L(f,\mathcal P_1)&=\sum_{i=1}^{N_1}(y_i-y_{i-1})(L_i-\ell_i)\\ &=\sum_{i=1}^{N_1}\frac{\delta}{\kappa(b-a)\lambda}(L_i-\ell_i)\end{align}$$

Now we split our set of indices $i=1,2,\dots,N_1$ into two subsets: $Q=\{i:\exists j\in[1,N_0]\text{ s.t. }[y_{i-1},y_i]\subseteq[x_{j-1},x_j]\}$ as the set of those indices where an interval of partition $\mathcal P_1$ is a subset of an interval of partition $\mathcal P_0$, and $R=\{1,2,\dots,N_1\}\setminus Q$ the set of indices where this is not the case. By our definition, each interval of $\mathcal P_1$ is no larger than any interval in $\mathcal P_0$, and therefore there are no cases of intervals of $\mathcal P_0$ being proper subsets of intervals of $\mathcal P_1$. Thus we can immediately state

$$\sum_{i\in Q}\frac{\delta}{\kappa(b-a)\lambda}(L_i-\ell_i)\le\sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)\lt\sigma_0$$

Note that there are $N_0-1$ cuts of $[a,b]$ associated with partition $\mathcal P_0$, and therefore we have $|R|\le N_0-1$. By straight evaluation, we have $L_i-\ell_i\le \lambda, {b-a\over \delta}\ge N_0$, so we get

$$\frac{\delta}{\kappa(b-a)\lambda}\sum_{i\in R}(L_i-\ell_i)\le \frac{N_0}{\kappa N_0}=\frac {1}{\kappa}\\ \sum_{i=1}^{N_1}(y_i-y_{i-1})(L_i-\ell_i)\le \sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)+\frac{1}{\kappa}\\ =\sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)+\frac{1}{\left[\frac 1{\sigma_0-U(f,\mathcal P_0)+L(f,\mathcal P_0)}\right]+1}\\ \lt \sum_{i=1}^{N_0}(x_i-x_{i-1})(M_i-m_i)+(\sigma_0-U(f,\mathcal P_0)+L(f,\mathcal P_0))= \sigma_0$$

This can be described as the "brute-force" approach, where we merely squeezed the set of "boundary values" into the space between the previous sum and the given $\sigma_0$, and let the non-boundary values do no worse than preserving the value of the previous sum.

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  • $\begingroup$ Finally got an effective partition count... $\endgroup$ – abiessu Apr 20 '15 at 18:15

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