Problem 4.1
If \(H\) and \(K\) are bounded sets and \(H \subseteq K\) then \(glb(K) \leq glb(H)\text{.}\)
We have already shown that if \(M\) is a bounded point set, then either \(M\) has a right-most point or there is a first point to the right of \(M\text{.}\) We shall call this number, whichever it is, the least upper bound of \(M\text{,}\) and we will denote it by lub(\(M\)). Similarly if a set \(M\) has a left-most point or a first point to the left of \(M\text{,}\) then we will refer to this point as the greatest lower bound of \(M\) and denote it by glb(\(M\)). Some mathematicians use the notation, supremum of \(M\) and infimum of \(M\) respectively.
If \(H\) and \(K\) are bounded sets and \(H \subseteq K\) then \(glb(K) \leq glb(H)\text{.}\)
If each of \(H\) and \(K\) are bounded sets, then \(H \oplus K = \{ h + k : h \in H, k \in K \}\text{.}\)
If each of \(H\) and \(K\) are bounded sets, then \(glb(H) + glb(K) = glb( H \oplus K )\text{.}\)
A bounded function is a function with bounded range.
If \([a,b]\) is a closed interval, by a partition of \([a,b]\) is meant a set of points \(\{t_0,t_1, \dots , t_n \}\) satisfying \(a=t_0 \lt t_1 \lt t_2 \lt \dots \lt t_{n-1} \lt t_n=b\text{.}\)
For the next four definitions, assume that \(f\) is a bounded function with domain the closed interval \([a,b]\text{.}\)
The statement that the number \(S\) is a Riemann sum for \(f\) on \([a,b]\) means that there is a partition \(\{t_0, t_1, \dots, t_n\}\) of \([a,b]\) and a sequence \(x_1,x_2,\dots,x_n\) of numbers such that \(\displaystyle x_i \in [t_{i-1},t_i] \mbox{ for } i=1,2,3,\dots,n\) and \(\displaystyle S = \sum_{i=1}^{n} f(x_i)(t_i-t_{i-1})\text{.}\)
The statement that the number \(S\) is the upper Riemann sum for \(f\) on \([a,b]\) means that there is a partition \(\{t_0, t_1, \dots, t_n\}\) of \([a,b]\) and a sequence \(y_1,y_2,\dots,y_n\) of numbers such that \(y_i = lub\{f(x)|x \in [t_{i-1},t_i]\}\) for \(i= 1,2,\dots,n\) and \(\displaystyle S = \sum_{i=1}^n {y_i (t_i - t_{i-1})}\text{.}\)
We define the lower Riemann sum in the same way except that \(y_i=glb\{f(x)|x \in [t_{i-1},t_i]\}\) for each positive integer \(i=1,2,\dots,n\text{.}\)
If \(f\) is a bounded function with domain the closed interval \([a,b]\) and \(P\) is a partition of \([a,b]\text{,}\) then \(U_P(f)\) and \(L_P(f)\) denote the upper and lower Riemann sums of \(f\text{.}\)
Let \(f(x)=0\) for each number \(x\) in \([0,1]\) except \(x=0\text{,}\) and let \(f(0)=1\text{.}\) Show that:
if \(P\) is a partition of \([0,1]\text{,}\) then \(0 \lt U_P f,\)
if \(\epsilon > 0,\) then there is a partition \(P\) of \([0,1]\) such that \(U_P f \lt \epsilon\text{,}\) and
zero is the only lower Riemann sum for \(f\) on \([0,1]\text{.}\)
Show that if \(f\) is a function whose domain includes the closed interval \([a,b]\text{,}\) and for each number \(x\) in \([a,b]\text{,}\) \(m\le f(x)\le M\text{,}\) and \(P=\{t_0,t_1,\dots,t_n\}\) is any partition of \([a,b]\text{,}\) then \(U_P f \le M (b-a)\) and \(L_P(f) \ge m (b-a)\text{.}\)
If \(f\) is continuous at the point \(p\) and \(K\) is a subset of the domain of \(f\) and \(p\) is a limit point of \(K,\) then \(f(p) \in Cl(f(K)),\) the closure of \(f(K)\text{.}\)
If \(f\) is bounded on \([a,b]\) then the set of all Riemann sums of \(f\) is bounded.
If \(f\) is a bounded function with domain the closed interval \([a,b],\) then the upper integral from \(a\) to \(b\) of \(f\) is the greatest lower bound of the set of all upper Riemann sums for \(f\) on \([a,b]\) and is denoted by \(\displaystyle _U\int_a^b f\text{.}\) The lower integral from \(a\) to \(b\) of \(f\) is the least upper bound of the set of all lower Riemann sums for \(f\) on \([a,b]\) and is denoted by \(\displaystyle _L\int_a^b f\text{.}\)
If \(f\) is a bounded function with domain \([a,b],\) then the statement that \(f\) is Riemann integrable on \([a,b]\) means that \(\displaystyle _L\int_a^b f=_U\int_a^b f\text{.}\)
When a function is Riemann integrable, we drop the subscripts \(U\) and \(L\) and refer to \(\displaystyle \int_a^b f\) as the Riemann integral of \(f\text{.}\)
The statement that the partition \(Q\) of the closed interval \([a,b]\) is a refinement of the partition \(P\) of \([a,b]\) means that \(P \subseteq Q\text{.}\)
If \(P_1\) and \(P_2\) are partitions of \([a,b]\) then there exists a partition \(Q\) of \([a,b]\) so that \(Q\) is a refinement of both \(P_1\) and \(P_2\text{.}\)
If \(f\) is a function with domain \([a,b]\text{,}\) and \(f\) is continuous at each number in \([a,b]\text{,}\) then the range of \(f\) is a closed point set.
A stronger statement is true: If \(f\) is continuous on the closed interval \([a,b]\) and \(M \subseteq [a,b]\) is closed then \(f(M)\) is closed.
If \(f\) is a bounded function with domain the closed interval \([a,b],\) \(P\) is a partition of \([a,b]\text{,}\) \(Q\) is a partition of \([a,b],\) and \(Q\) is a refinement of \(P\text{,}\) then \(L_P(f) \le L_Q(f)\) and \(U_P(f) \ge U_Q(f)\text{.}\)
If each of \(H\) and \(K\) are sets and for all \(h \in H\) and \(k \in K\) we have \(h \lt k\text{,}\) then \(lub(H) \leq glb(K)\text{.}\)
If \(f\) is a bounded function with domain \([a,b],\) then \(\displaystyle _L\int_a^b f \; \leq \; _U\int_a^b f\text{.}\)
If \(f\) is a bounded function with domain \([a,b]\text{,}\) and for each number \(x\) in \([a,b]\text{,}\) \(f(x) \ge 0\text{,}\) and for some number \(z\) in \([a,b]\text{,}\) \(f(z) > 0\) and \(f\) is continuous at \(z\text{,}\) then \(\displaystyle _U \int_a^b f>0\text{.}\)
If \(f\) is a continuous function with domain the closed interval \([a,b]\text{,}\) and \(\epsilon\) is a positive number, then there is a partition \(\{ x_0, x_1,x_2 , \dots , x_n \}\) of the closed interval \([a,b]\) such that for each positive integer \(i\) not larger than \(n\text{,}\) if \(u\) and \(v\) are two numbers in the closed interval \([x_{i-1},x_i]\text{,}\) then \(|f(u)-f(v)| \le \epsilon\text{.}\)
If \(f\) is a bounded function with domain the closed interval \([a,b]\) and for each positive number \(\epsilon\text{,}\) there is a partition \(P\) of \([a,b]\) such that \(U_P(f)-L_P(f)\lt \epsilon\text{,}\) then \(f\) is Riemann integrable on \([a,b]\text{.}\)
If \(f\) is a continuous function with domain the closed interval \([a,b],\) then \(f\) is Riemann integrable on \([a,b]\text{.}\)
A function \(f\) is increasing if for each pair of points \(x\) and \(y\) in the domain of \(f\) satisfying \(x\lt y\) we have \(f(x) \lt f(y)\text{.}\) The function is non-decreasing if under the same assumptions we have \(f(x) \leq f(y)\text{.}\)
Every non-decreasing bounded function on \([a,b]\) is Riemann integrable on \([a,b]\text{.}\)
If \(f\) is a function with domain the closed interval \([a,b]\) and there is a sequence \(x_1,x_2 , \dots\) of points in \([a,b]\) converging to the point \(c\) and for each positive integer \(n,\) we have that \(f(x_n)=n\text{,}\) then \(f\) is not continuous at \((c,f(c))\text{.}\)
If \(f\) is a bounded function with domain the closed interval \([a,b]\text{,}\) and \(P\) is a partition of \([a,b]\text{,}\) then \(L_ P (f) \le U_P (f)\text{.}\)
If \([a,b]\) is a closed interval and \(c \in (a,b)\) and \(f\) is integrable on \([a,c]\) and on \([c,b]\) and on \([a,b]\text{,}\) then \(\displaystyle \int_a^c f + \int_c^b f=\int_a^b f\text{.}\)
If \([a,b]\) is a closed interval and \(f\) is integrable on \([a,b]\) then we define \(\displaystyle \int_b^a f \; = \; -\int_a^b f\) and \(\displaystyle \int_a^a f = 0\text{.}\)
If \(f\) is a continuous function with domain the closed interval \([a,b]\text{,}\) then there is a number \(c\) in \([a,b]\) such that \(\displaystyle \int_a^b f = f(c) (b-a)\text{.}\)
If \(f\) is a continuous function with domain \([a,b]\)>, then the range of \(f\) is bounded.
If \(f\) is a continuous function with domain \([a,b]\text{,}\) then there is a number \(x \in [a,b]\) such that if \(t \in [a,b]\text{,}\) then \(f(t) \le f(x)\text{.}\)
If \(f\) is a continuous function with domain \([a,b]\text{,}\) then the range of \(f\) contains only one value or it is a closed interval.
Suppose \(f\) is a function whose domain includes \([a,b],\) \(f(a) = 0 = f(b),\) and \(f\) has a derivative at each of its points. Then there is a number \(c\) in (a,b) such that \(f'(c)=0\text{.}\)
Suppose \(f\) is a function whose domain includes \([a,b]\) and \(f\) has a derivative at each of its points. Let \(L\) be the line between \((a,f(a))\) and \((b,f(b))\text{.}\) Show that the function \(g\) defined by \(g(x) = f(x) - L(x)\) satisfies the hypothesis of Lemma 4.35.
If \(f\) is continuous on \([a,b]\) and \(\displaystyle g(x) = \int_a^x f\) for all \(x \in [a,b]\) then \(g\) is continuous on \([a,b]\text{.}\)
If \(f\) is an integrable function, then \(\displaystyle | \int_a^b f | \leq \int_a^b |f|\text{.}\)
Suppose \(f\) is a non-decreasing function whose domain includes \([a,b]\) and \(f\) has a derivative at each of its points, then there is a number \(c\) in \((a,b)\) such that \(f'(c)\) is the same as the slope of the line joining the two points \((a,f(a))\) and \((b,f(b))\text{.}\)
Although we stated the previous theorem only for non-decreasing functions, it is valid for any differentiable function defined on \([a,b]\) and differentiable on \((a,b)\text{.}\) You may use the more general statement if you require it later.
If \(f\) is a continuous function with domain the closed interval \([a,b]\) and \(F\) is the function such that for each number \(x\) in \([a,b]\text{,}\) \(\displaystyle F(x)=\int_a^x f\text{,}\) then for each number \(c\) in \([a,b]\) \(F\) has a derivative at \(c\) and \(F'(c)~=~f(c)\text{.}\)
Show that if each of \(f\) and \(g\) are functions defined and differentiable on the closed interval \([a,b]\) and \(f'(x) = g'(x)\) for all \(x \in [a,b]\text{,}\) then there is a number \(k \in \R\) such that \(f(x) = g(x) + k\) for all \(x \in [a,b]\text{.}\)
If \(f\) is a function with domain the closed interval \([a,b]\) and \(f\) has a derivative at each point of \([a,b]\) and \(f'\) is continuous at each point in \([a,b]\text{,}\) then \(\displaystyle \int_a^b f' = f(b)-f(a)\text{.}\)