On the fundamental theorem of calculus
In [1], we defined the Riemann integral on intervals of \mathbb{R} . We shall now prove some theorems pertaining to it. Below, we will let S_f(P) denote the Riemann sum of f associated with tagged partition P.
Definition 1 A piecewise function f defined on a bounded interval I is defined by partitioning I into a finite number of sub-intervals I_1,I_2, \ldots, I_n and defining on each of the I_i s a function f_i such that for x \in I_i , f(x) = f_i(x) . We say that f is piecewise continuous if each of the f_i s is continuous. One can, here, replace "continuous" with any other qualification of function, such as "smooth".
Proposition 1 We have the following properties of Riemann integrable functions.
- If f is Riemann integrable on two disjoint intervals I_1,I_2 , with the Riemann integrals equal to \int_{I_1} f\;dx, \int_{I_2} f\;dx , then f is Riemann integrable on I_1 \cup I_2 , with\int_{I_1 \cup I_2} f\; dx = \int_{I_1} f\;dx+\int_{I_2} f\;dx.\\
- Let I_2 be an interval containing interval I_1 . If f is Riemann integrable on I_2 , it is Riemann integrable on both I_1 and I_2 \setminus I_1 , with \int_{I_2\setminus I_1} f\;dx = \int_{I_2} f\;dx - \int_{I_1} f\;dx.\\
- If on any interval I , both f and g are Riemann integrable, then f+g is also Riemann integrable on I , with \int_{I} f+g\;dx = \int_{I}f\;dx +\int_{I}g\;dx . Moreover for any c \in \mathbb{R} , cf is Riemann integrable on I with \int_I cf\;dx = c\int_If\;dx . In other word, the Riemann integral on I operator is a linear operator on the space of Riemann integrable functions on I .
Proof: For (1), we take a tagged partition P_1 of I_1 satisfying \epsilon/2>0 and a tagged partition P_2 of I_2 satisfying \epsilon/2>0 . Formally, this says that any P refinement of P_1 is such that \left|S_f(P_1) - \int_{I_1} f\;dx\right|< \epsilon/2 and similar for P_2 . For I_1 \cup I_2 we use the tagged partition P_1 \cup P_2 . One easily verifies with the triangle inequality that any refinement P of it satisfies \left|S_f(P) - \int_{I_1 \cup I_2} f\;dx\right|< \epsilon .
For (2), take any tagged partition P_2 of I_2 that contains the endpoints of I_1 . The equivalence of (3) of Lemma 3 from [2] with Riemann integrability tells us that for any \epsilon > 0 , there exists a refinement P_\epsilon of P_2 such that for any Q_1,Q_2 which are refinements of P_\epsilon ,
|S_f(Q_1)-S_f(Q_2)| < \epsilon. \qquad (0)\\ Both of Q_1,Q_2 can be split into tagged partitions of I_1 and I_2\setminus I_1 , which we denote via Q_{11}, Q_{12} and Q_{22},Q_{21} (where the second index represents whether or not the partition is defined on I_1 or I_2\setminus I_1 , such that
S_f(Q_1) = S_f(Q_{11})+S_f(Q_{12}),\\ S_f(Q_2) = S_f(Q_{21})+S_f(Q_{22}). If by contradiction f were not Riemann integrable on one of I_1, I_2\setminus I_1 , then there would by Lemma 3 of [2] exist an \epsilon_1>0 for which there does not exist a partition P_{\epsilon_1} that is a refinement of P_2 such that the split of any two refinements of it into partitions on I_1, I_2\setminus I_1 , which we denote as Q_{11}',Q_{12}' and Q_{21}', Q_{22}' , satisfies both
|S_f(Q_{11}')-S_f(Q_{21}')| < \epsilon_1,\\ |S_f(Q_{12}')-S_f(Q_{22}')| < \epsilon_1. Setting \epsilon = \epsilon_1 and using the triangle inequality on the above would thus violate the existence of a refinement P_\epsilon of P_2 which guarantees (0) , which would imply that f is not Riemann integrable on I_2 , a contradiction.
With similar logic, one proves Riemann integrability of f on I_2\setminus I_1 . The integral equality of (2) then follows directly from (1) .
The proof of (3) we leave as an exercise to the reader. \square
Proposition 2 Any continuous function on [a,b] \subset \mathbb{R} is Riemann integrable.
Proof: We construct a sequence of tagged partitions of [a,b] . We let the k th tagged partition be given by
a < a + \frac{1}{2^k}(b-a) < a + \frac{2}{2^k}(b-a) < \ldots < a + \frac{2^k-1}{2^k}(b-a) < b,\\
in which we have let [x_j, x_{j+1}] = \left[a + \frac{j}{2^k}(b-a), a + \frac{j+1}{2^k}(b-a)\right] be the subintervals of [a,b] corresponding to the partition, where j =0,1,\ldots, 2^k-1 . As for the tags, we let t_j = \inf_{t \in [x_j, x_{j+1}]} \{t : f(t) \leq \inf_{x \in [x_j, x_{j+1}]}f(x)\} , noting that by the Extreme Value Theorem, the infimum here exists. We notice that under the partial order of partition refinement, this sequence of tagged partitions is a chain. Moreover, the Riemann sums of these partitions form a bounded (by Extreme Value Theorem) monotonically non-decreasing sequence. Thus, by the Monotone Convergence Theorem, this sequence of Riemann sums converges to some L \in \mathbb{R} . For any \epsilon>0 , we can find a partition from this sequence such that the distance between its Riemann sum and L is less than \epsilon . Since a refinement of a tagged partition where the tags correspond to infimum of f on subintervals cannot decrease the Riemann sum, L satisfies the definition of Riemann integral. We emphasize in this proof that the subintervals in the definition of Riemann sum are closed and are not disjoint at endpoints, which allows for two tags (of different subintervals) to have the same value. \square
Corollary 1 Any function f continuous and bounded on an open or half-open interval in \mathbb{R} is Riemann integrable. Moreover, if we extend such a function to the closure of that interval via the limit and Riemann integrate this on the closure, the result is the same. Formally, \int_a^b f\;dx \equiv \int_{(a,b)} f\;dx = \int_{[a,b)} f\;dx = \int_{(a,b]} f\;dx = \int_{[a,b]} f\;dx . Because of this, we can disregard whether or not the endpoints are included in the interval itself and simply use \int_a^b to denote this Riemann integral.
Proof: We note that [a,a] and [b,b] are both closed intervals on which the Riemann integral exists and is zero. The Riemann integral exists on [a,b] by Proposition 2. Applying (2) of Proposition 1 tells us that the Riemann integral exists and is equal to \int_{[a,b]} f\;dx on each of (a,b), (a,b], [a,b) . \square
Definition 2 Suppose that a < b . Then, for any Riemann integrable f on [a,b] we define \int_b^a f(x)dx = -\int_a^b f(x)dx.
Proposition 3 Any function f piecewise continuous on any interval is Riemann integrable on that interval.
Proof: Corollary 1 tells that we can WLOG only consider closed intervals. Let the interval be [a,b] , with [a,b] = \bigcup_{i=1}^n I_i , where the I_i s are disjoint intervals on each of which f is continuous. Proposition 2 tells us that f is Riemann integrable on each of the I_i s. Thus, we can apply (1) of Proposition 1 to derive that f is Riemann integrable on [a,b] .
One can also use the equivalence of (2) and (3) in Theorem 1 of [2], of which this is a very special case. \square
Proposition 4 On some interval [a,b] , define Riemann integrable f,g such thatf(x) < g(x) for all x \in [a,b] . Then,
\int_a^b f(x)dx < \int_a^b g(x)dx.\\
Proof: Trivial and left to the reader.
Theorem 1 (Mean value theorem for definite integrals) Let f: [a,b] \in \mathbb{R} be a continuous function. Then, there exists c \in [a,b] such that \int_a^b f(x)dx = (b-a)f(c).\\ Proof: The Extreme Value Theorem gives us infimum and supremum values of f on [a,b] , which we denote with L and M . This gives us
L(b-a) \leq \int_a^b f(x)dx \leq M(b-a).\\ By the Intermediate Value Theorem, for every point in y \in [L,M] is such that there exists c \in [a,b] such that f(c) = y . Applying this to the value of \frac{1}{b-a}\int_a^b f(x)dx yields the desired result. \square
Theorem 2 (Fundamental theorem of calculus, part I) Let f be a continuous real-valued function defined on closed interval [a,b] . Let F be the function defined, for all x \in [a,b] , by
F(x) = \int_a^x f(t)dt.\\ Then F is uniformly continuous on [a,b] and differentiable on the open interval (a,b) , and
F'(x) = f(x)\\ for all x \in (a,b) .
Proof: Take arbitrary x \in (a,b) . We can define for all h>0 such that [x-h,x+h] \subset [a,b]
\begin{eqnarray} F(x+h) - F(x) &=& \int_a^{x+h} f(t)dt - \int_a^x f(t)dt,\\ &=& \int_x^{x+h} f(t)dt. \end{eqnarray}\\ Applying Mean Value Theorem for Definite Integrals (Theorem 1) to this gives some c_h \in [x, x+h] such that
\frac{F(x+h) - F(x)}{h} = f(c_h).\\
As h \to 0 , c_h \to x . Because f is continuous on [x,x+h] , we thus have
F'(x) = \lim_{h\to 0}\frac{F(x+h) - F(x)}{h} = f(x).\\
By the Extreme Value Theorem, there exists M such that |F'(x)| = |f(x)| < M for all x \in [a,b] . Thus, by Proposition 4, for any \epsilon > 0 , take \delta = \epsilon / M and we have F((x-\delta, x+\delta) \cap [a,b]) \subset (F(x)-\epsilon, F(x)+\epsilon) for all x \in [a,b] . This shows uniform continuity. \square
Definition 3 We say that F is an antiderivative of f on [a,b] if on (a,b) , F'(x) = f(x) .
Lemma 1 Any continuous antiderivative F of the zero function on [a,b] is a constant function on [a,b].
Proof: Suppose by contradiction that some F that is a continuous anti-derivative of the zero function on [a,b] is not constant. Then there exists x_1, x_2 \in [a,b] such that F(x_1) < F(x_2) . The mean value theorem for differentiable functions tells us that there exists some c \in (x_1, x_2) such that F'(c) = \frac{F(x_2) - F(x_1)}{x_2-x_1} \neq 0 , which means that F would then not be a continuous function on [a,b] with derivative equal to zero on (a,b) . \square
Lemma 2 If a function f defined on [a,b] has antiderivatives F_1, F_2 on (a,b) , then F_1-F_2 is a constant function on [a,b] , where the values of F_1,F_2 at a,b are simply defined via continuous extension.
Proof: The Fundamental Theorem of Calculus, Part I (Theorem 1) tells us that
F(x) = \int_a^x f(t)dt\\ is an antiderivative of f on (a,b) that is continuous on [a,b] . Because the relation defined via constant difference is transitive, it suffices to simply show that F_1(x) - F(x) is constant on [a,b] .
That F_1'(x) = F'(x) = f(x) on (a,b) tells us that F_1(x) - F(x) is an antiderivative of the zero function on (a,b) , from which we deduce via Lemma 1 that is a constant function. \square
Corollary 2 If f is a real valued continuous function on [a,b] and F is a continuous antiderivative of f on [a,b] , then
\int_a^b f(t)dt = F(b) - F(a).\\
Proof: We define F_1(x) = \int_a^x f(t)dt on [a,b] . Fundamental Theorem of Calculus Part I tells us that it is a continuous antiderivative of f on [a,b] . We have that
F_1(b) - F_1(a) = \int_a^b f(t)dt.\\
By Lemma 2, F_1 - F is a constant function on [a,b] , we must have that F(b) - F(a) = F_1(b) - F_1(a) , which completes our proof. \square
Theorem 2 (Fundamental theorem of calculus part II (Newton-Leibniz axiom)) If f is a real valued function on [a,b] and Riemann integrable on [a,b] and F is an antiderivative of f on [a,b] then
\int_a^b f(t)dt = F(b) - F(a).\\
Proof: Take an arbitrary partition a = x_0 < x_1 < \ldots < x_n = b . We have that
F(b) - F(a) = \sum_{i=0}^{n-1} [F(x_{i+1}) - F(x_i)].\\ We can apply the mean value theorem for differentiability on the subintervals [x_i, x_{i+1}] to assign tags such that F(x_{i+1}) - F(x_i) = (x_{i+1} - x_i)f(t_i) which yields the Riemann sum
F(b) - F(a) = \sum_{i=0}^{n-1} [(x_{i+1} - x_i)f(t_i)]. \qquad (3)\\
By definition of Riemann integrable as given in (4) of Theorem 1 of [2], for all \epsilon >0 , there exists \delta > 0 such that for any tagged partition P of [a,b] denoted by a = x_0 < x_1 < \ldots < x_{n-1} < x_n = b and t_0,t_1,\ldots, t_{j-1} for the tags such that its norm or mesh \parallel P\parallel = \max_{i=0,1,\ldots,n-1} (x_{i+1}-x_j) < \delta , we have
\left|S_f(P) - \int_a^b f(t)dt\right| < \epsilon.\\
Another way to put it, per Lemma 2, is that for all \epsilon > 0 , there exists \delta > 0 such that any tagged partitions P_1,P_2 with meshes both less than \delta ,
\max(\parallel P_1 \parallel,\parallel P_2 \parallel)<\delta,\\
such that
|S_f(P_1) - S_f(P_2)| < \epsilon.\\
Thus, we can, setting \parallel x\parallel = \max_{i=1}^{n-1}(x_{i+1}-x_i) , take the limit of (3) to derive
\begin{eqnarray} F(b) - F(a) &=& \lim_{\parallel x\parallel \to 0}(F(b) - F(a)),\\ & =& \lim_{\parallel x\parallel \to 0}\sum_{i=0}^{n-1} [(x_{i+1} - x_i)f(t_i)],\\ &=& \int_a^bf(t)dt, \end{eqnarray}\\ which completes our proof. \square
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