So I've been using the term "Riemann Integrable" a few times, but what does it mean to be riemann integrable on a closed interval $[a,b]$.
The Riemann integral is a method of exhaustion applied to finding the area under the curve. The area is approximated using rectangles with decreasing widths. As an infinite number of rectangles are used to find the area this ceases to be an approximation and the true area is found. I talked about this in my previous post but it is important to understand the intuition behind the riemann integral.
First we will consider the function $f: [a,b] \mapsto \mathbb{R}$
As we did previously we want to split up the interval $[a,b]$, where $a < b$. In this case we split the interval into a set of partitions $\mathcal{P}$ that we define as: \[ \mathcal{P} := \{ a = x_0 < x_1 < ... < x_{n-1} < x_n = b \} \] We will also define the norm of this partition \[ || \mathcal{P} || := \max_{1 \leq i \leq n} \Delta x_i\] where we define the partition width as (the partitions may be of different sizes rather than uniform as we were dealing with before): \[\Delta x_i = x_i - x_{i-1}\] We're going to define another partition on $[a,b]$ and call it $\mathcal{P}'$, this new partition $\mathcal{P}'$ is a refinement of $\mathcal{P}$. This is done by inserting additional points between those defined by $\mathcal{P}$ such that $ \mathcal{P} \subseteq \mathcal{P}'$. Now if $f$ is defined on [a,b] we can write the sum \[\Lambda (f, \mathcal{P})= \sum_{i=1}^{n} f(x) \Delta x_i \] where $x \in [x_{i-1} , x_i]$ and $1 \leq i \leq n$. $\Lambda$ is the riemann sum of $f$ over the partition $\mathcal{P}$. It is important to note that there are infinitely many riemann sums of a single function $f$ over a given partition $\mathcal{P}$ as over the interval $[x_{i-1},x_i]$ the reals are infinitely dense and $x$ can take any real number in the interval as its value.
Now we're going to define what it means to be riemann integrable, we say that a function $f$ is riemann integrable on $[a,b]$ if there exists a unique number $\mathcal{L} \in \mathbb{R}$ with the property that for all $\epsilon > 0$ there is a $\delta > 0 $ such that \[ | \Lambda - \mathcal{L} | < \epsilon \] Where $\Lambda$ is an arbitrary riemann sum of $f$ over the partition $\mathcal{P}$ of $[a,b]$ such that $|| \mathcal{P} || < \delta $. If we take the limit as $|| \mathcal{P} || \to 0$ We can say that $\mathcal{L}$ is the riemann integral of $f$ on $[a,b]$ and denote this in the familiar way as: \[ \int_{a}^{b} f(x) dx = \mathcal{L} \] I'm not going to prove here that $\mathcal{L}$ is unique, however this result will soon become apparent.
If you're familiar with real analysis you way want to look away for this paragraph as the following "definition" may offend you. Now I'm going to introduce some new notation that applies to sets, the supremum, $\sup$ and infimum, $\inf$. Loosely speaking these correspond to maximum and minimum elements of an arbitrary set $\mathcal{S}$.
Now we want to define the maximum value of $f(x)$ over an arbitrary partition $[x_{i-1},x_i]$ and denote this as $J_i$ \[ J_i := \sup \ \{ \ f(x) : \ {x \in [x_{i-1}, x_i]} \ \} \] Now we want to do the same with the lower partition denoting it as $j_i$ \[ j_i := \inf \ \{ \ f(x) : \ {x \in [x_{i-1}, x_i]} \ \} \]
From these we can now form the upper and lower riemann sums, these flavours of sum represent the maximum and minimum area under the curve $f(x)$ over the partition $[a,b]$.
First the upper sum of $f$ over the partition $\mathcal{P}$ \[ U(f,\mathcal{P}) := \sum_{i=1}^{n} J_i \Delta x_i\] And the lower sum of $f$ over $\mathcal{P}$ \[ L(f,\mathcal{P}) := \sum_{i=1}^{n} j_i \Delta x_i\]
Now let $f$ be bounded in the interval $[a,b]$ such that $j \leq f \leq J$, where $j, J \in \mathbb{R}$ and are defined as follows: $j = \inf \ \{ \ f(x) : \ x \in [a,b] \ \}$ and $J = \sup \ \{ \ f(x) : \ x \in [a,b] \ \}$. Now using our definition of the upper and lower riemann sums we can say that \[ j(b-a) \leq L(f,\mathcal{P}) \leq U(f,\mathcal{P}) \leq J(b-a) \] Now we return to our refined partition $\mathcal{P}'$, suppose it refines the partition $\mathcal{P}$ by adding in a single extra point $r$ such that $\mathcal{P}' = \mathcal{P} \cup \{r\}$. Now suppose that there exists an integer $k$ such that $x_{k-1} \leq r \leq x_k$. Now suppose that this new point $r$ divides the $k^{th}$ term, this means the final interval is now divided into two intervals that we shall call the left and right interval. \[ j_i^L := \inf \ \{ \ f(x) : \ x \in [x_{i-1}, r] \ \} , \ \ \ \ J_i^L := \sup \ \{ \ f(x) : \ x \in [r, x_i]: \ f(x) \ \} \] \[ j_i^R := \inf \ \{ \ f(x): \ x \in [x_{i-1}, r] \ \} , \ \ \ \ J_i^R := \sup \ \{ \ f(x) : \ x \in [r, x_i] \ \} \] Now we can calculate the riemann sums of the refined partition, note that as only the last term is changed by the additional point so we only need to calculate this \[ J_i \Delta x_i = J_i^{L}(r - x_{i-1}) + J_i^{R} (x_i - r)\] \[ j_i \Delta x_i = j_i^{L}(r - x_{i-1}) + j_i^{R} (x_i - r)\] From this it follows that \[ U(f,\mathcal{P}) - U(f,\mathcal{P}') = (J_i - J_i^L)(r-x_{i-1}) + (J_i - J_i^R)(x_i - r) \] \[ L(f,\mathcal{P}) - L(f,\mathcal{P}') = (j_i - j_i^L)(r-x_{i-1}) + (j_i - j_i^R)(x_i - r) \] Now it may take a little bit of thought but by following our definitions for the right and left intervals we can note that for the left interval $j_i \leq j_i^L \leq J_i^L \leq J_i$ and that $j_i \leq j_i^R \leq J_i^R \leq J_i$ analogously for the right interval. From this it follows that \[ L(f,\mathcal{P}) \leq L(f,\mathcal{P}') \leq U(f,\mathcal{P}') \leq U(f,\mathcal{P}) \] This is what we're looking for, hold this result as we will use it later. In summary this basically says that if we refine some partition $\mathcal{P}$ by adding in additional points the lower riemann sum will increase and the upper riemann sum will decrease.
Now let $\mathcal{P}_1$ and $\mathcal{P}_2$ be two partitions on $[a,b]$. Now if we let them both be a refinement on $\mathcal{P}'$ such that $ \mathcal{P}' := \mathcal{P}_1 \cup \mathcal{P}_2$. We can now use theorem 1.3 to establish the inequality \[ L(f,\mathcal{P}_1) \leq L(f,\mathcal{P}') \leq U(f,\mathcal{P}') \leq U(f,\mathcal{P}_2) \] This is a very important results as it means the the lower sum can never exceed the upper sum regardless of how we choose the partitions of summation. Which leads us to \[ \sup \ \{ \ L(f,\mathcal{P}) : \ \mathcal{P} \in [a,b] \ \} \leq \inf \ \{ \ U(f,\mathcal{P}) : \ \mathcal{P} \in [a,b] \ \} \] Which we will need to define the riemann integral
We haven't quite finished here yet earlier when we added in points we only considered adding in one additional interval into the partition $\mathcal{P}$. Now lets consider the more general case when we add $N$ more points. I'm not going to prove it, but from theorem 1.3 it follows that if we refine the partition by adding another point the inequality still holds. So we can add $N$ additional points and the inequality will still hold.
As you may have guessed we're now going to form riemann integrals from the summations we have constructed. We're in a position to define the upper and lower riemann sums that correspond to these summations.
The upper riemann integral is defined as: \[ \overline{\int_{a}^{b}} f(x) dx := \inf \ \{ \ \mathcal{P} \in [a,b]: \ \ U(f,\mathcal{P}) \ \} \] The lower riemann integral is defined as \[ \underline{\int_{a}^{b}} f(x) dx := \sup \ \{ \ \mathcal{P} \in [a,b]: \ \ L(f,\mathcal{P}) \ \} \] So we can interpret the upper riemann integral as the lowest upper bound of $f$, this is because we can vary the partition size to change the value of $U$, utilising the infimum and supremum to do this. Likewise we can do the same to the lower riemann sum $L$ except in this case $f$ is the greatest lower bound. \[ \underline{\int_{a}^{b}} f(x) dx \leq \overline{\int_{a}^{b}} f(x) dx \] This inequality is formed by using theorem 1.4 And from it we arrive at one of the conditions for a function $f$ to be riemann integrable on $[a,b]$. The function $f$ is integrable on $[a,b]$ if and only if the upper and lower sums converge on a common value and we denote this as $\int_a^b f(x) dx$.
From definition 1.6 a function $f$ is riemann integrable on $[a,b]$ if and only if a unique limit exists which means \[ \overline{\int_{a}^{b}} f(x) dx = \underline{\int_{a}^{b}} f(x) dx = \int_{a}^{b} f(x) dx \] Any integrable function $f$ will fulfil it. We can reformulate this expression to form the Riemann lemma, a condition for integrability that will be very useful in identifying riemann integrable functions.
A function $f : [a,b] \mapsto \mathbb{R}$ is riemann integrable on $[a,b]$ if and only if for a partition $\mathcal{P}$ on $[a,b]$ for all $\epsilon > 0$ such that \[ \Big| U(f, \mathcal{P}) - L(f, \mathcal{P}) \Big| < \epsilon \] To prove this suppose that there is a real number $\mathcal{L}$ and an $\epsilon > 0$ and consider the interval $\Big[ \mathcal{L}, \mathcal{L} + \dfrac{\epsilon}{2} \Big]$ Clearly all the lower sums are less than or equal to $\mathcal{L}$ and the upper sums are greater than or equal to $\mathcal{L}$. So for some partition $\mathcal{P}_2$ we have \[ \mathcal{L} \leq U(f, \mathcal{P}_2) \leq \mathcal{L} + \dfrac{\epsilon}{2} \] Now a similar inequality exists for the lower sum. Consider this over the partition $\mathcal{P}_1$ \[ \mathcal{L} - \dfrac{\epsilon}{2} \leq L(f, \mathcal{P}_1) \leq \mathcal{L} \] Now consider the refined partition $\mathcal{P}' := \mathcal{P}_1 \cup \mathcal{P}_2 $ Now combining these together into a single inequality which, unsurprisingly, resembles theorem 1.3 & 1.4 \[ \mathcal{L} - \dfrac{\epsilon}{2} \leq L(f,\mathcal{P}_1) \leq L(f,\mathcal{P}') \leq U(f,\mathcal{P}') \leq U(f,\mathcal{P}_2) \leq \mathcal{L} + \dfrac{\epsilon}{2} \] Both $U(f, \mathcal{P}')$ and $L(f, \mathcal{P}')$ lie within the boundary \[ \Big[ \mathcal{L} - \dfrac{\epsilon}{2}, \mathcal{L} + \dfrac{\epsilon}{2} \Big] \] And the Riemann lemma follows from this.
So a function $f$ for to be integrable it must fulfil the Riemann lemma and if it fulfils this criterion we say $f$ is riemann integrable on $[a,b]$ and we denote this value as \[ \int_{a}^{b} f(x) dx = \mathcal{L} \]
No comments:
Post a Comment