Proof tips

• integrability: show that $\lim U(f, P_n) - L(f, P_n) = 0$ for some sequence $P_n$.
  • Equivalent: show that for every $\epsilon$ ,there exists some $P_\epsilon$ that satisifes $U(f, P_\epsilon) - L(f, P_\epsilon) < \epsilon$. This typically means finding some sort of rule for the partition, like distance.
  • it’s often convenient to show this difference

• if something is integrable, you can use the existence of $P_\epsilon$ for every $\epsilon$.

• smaller minus larger means that the overall difference is smaller.

• look for telescoping stuff in Darboux sums

• Use bounded fact in the sum

• Range of a subinterval is always at most the range of the larger interval

Darboux Integrals

Partitions 🐧

To establish the Darboux integral, we first need to talk about a partition, $P$. We define a partition on $[a, b]$ as a set

We call $Q$ a subpartition of $P$ if $P \subseteq Q$. Intuitively, $Q$ has more slices than $P$.

Intuitively, we also know that if $P, Q$ are both partitions, then $R = P \cup Q$ is also a partition.

Darboux Sum 🐧

let $M(f, S) = \sup \{f(x)\}$ and $m(f, S) = \inf\{f(x)\}$ where $S$ is one segment of a partition, $[x_{k-1}, x_k]$.

Using this notation, we can define the upper Darboux sum as

and the lower Darboux sum as

Properties of Darboux Sums 🚀

Obviously, from our definition, we know that $U(f, P) \geq L(f, P)$ because each $M \geq m$.

We can also say this: If $Q$ is a subpartition of $P$, then $U(f, Q) \leq U(f, P)$ and $L(f, Q) \geq L(f, Q)$.

• Proof (using definitions of Darboux sums)

  Let’s just say for simplicity that $Q$ only has one more partition than $P$. If we can show that this holds, we can easily show (through induction) that it’s true for any subpartition. We will also only show it for $U$; the $L$ follows easily.

  Now, suppose that $Q$ has intervals $[x_t, k], [k, x_{t+1}]$ while $P$ just has $[x_t, x_{t+1}]$. Without loss of generality, let’s assume that the maximum of $[x_t, x_{t+1}]$, $M$, falls into $[x_t, k]$. Therefore, the maximum of $[x_t, k]$ must be $M$. Furthermore, the maximum of $[k, x_{t+1}]$ must be less than $M$. Let’s denote this with $M’$.

  So, let’s consider the Darboux sum in this interval. For $Q$, we have $M * (k - x_t) + M’ * (x_{t+1} - k)$, while for $P$, we have $M * (k - x_t) + M * (x_{t+1} - k)$. It is clear that this sum for $P$ is larger than for $Q$.

  By simple induction, we see that this works when $Q$ has more than one partition than $P$.

The Darboux Integral 🐧

We propose the following definition for a Darboux integral:

For notational purposes, we call $U(f) = \inf\{ U(f, P)\}$, where this means the greatest lower bound of this value across all possible partitions. Similarly, we use the shorthand $L(f) = \sup\{L(f, P)\}$

Naturally, if $U(f) \neq L(f)$, then the function is not integrable.

• Examples

  To integrate $f(x) = 1$, we consider any arbitrary partition. We have

  $$U(f, P) =\sum_k M(f[x_{k-1}, x_k])(x_k - x_{k-1})$$

  and we know that this $M = 1$, so the summation collapses into a telescoping sequence, yielding $U(f, P) = b -a$. Similarly, we get that $L(f, P) = b - a$. Therefore, $U(f) = L(f) = b - a$.

  Let’s try to integrate the exotic function $f(x) = 1$ if $x \in \mathbb{Q}$, and $f(x) = 0$ if $x \notin \mathbb{Q}$. Let’s set this up:

  $$U(f, P) =\sum_k M(f[x_{k-1}, x_k])(x_k - x_{k-1})$$

  By the density of rational and irrational numbers, we know that $M = 1$ for any partition, so $U(f, P) = b - a$. However, consider

  $$L(f, P) =\sum_k m(f[x_{k-1}, x_k])(x_k - x_{k-1})$$

  We actually know that $m = 0$ regardless of partition, so we get $L(f, P) = 0$. Therefore, $U(f) = 1, L(f) = 0$, and $1 \neq 0$ so the function is not integratable.

Darboux Integrability Theorem 🚀 ⭐

Theorem: A bounded function on $[a, b]$ is Darboux-Integrable if and only if there exists a sequence of partitions $P_n$ such that $\lim_{n→\infty} U(f, P_n) = \lim_{n → \infty} L(f, P_n)$.

• Proof

  Let’s use the definition of Darboux sums. Because $U(f) = \inf U(f, P)$, we know that there must exist some sequence $\{P_n\}$ such that $\lim U(f, P_n) = U(f)$. Why? Well, you can construct it using the definition of an infimum. If you have $P_k$ such that $U(f, P_k) > U(f)$, then there must be some $P_{k+1}$ such that $U(f, P_{k+1}) < U(f, P_k)$ or else you contradict the infimum definition.

  Similarly, you can find some sequence $\{Q_n\}$ such that $\lim L(f, Q_n) = L(f)$. Note that we are not using any Darboux integrability, as we need to show that in our proof; this is just the Darboux sums.

  Now…consider the partition $R_n = Q_n \cup P_n$. By partition theorems, we know that $R_n$ is a partition. What’s more, we have $U(f, R_n) \leq U(f, P_n)$ and $L(f, R_n) \geq L(f, Q_n)$.

  We are actually really close to finishing this proof.

  Let’s consider the forward direction. If $f$ is Darboux integrable, then we have $U(f) = L(f)$. We know that $\lim U(f, R_n) \leq \lim U(f, P_n) = U(f)$. We also know that $\lim L(f, R_n) \geq \lim L(f, Q_n) = L(f)$. Therefore, we have $\lim L(f, R_n)\geq \lim U(f, R_n)$. However, we also know by construction of $L, U$ that $\lim L(f, R_n) \leq \lim U(f, R_n)$ for any sequence. Therefore, we must conclude that $\lim L(f, R_n) = \lim U(f, R_n)$. This is the sequence of partitions!

  Let’s consider the backward direction. If we have $\lim_{n→\infty} U(f, P_n) = \lim_{n → \infty} L(f, P_n)$ then for all $\epsilon > 0$, there must exist some $N_\epsilon$ such that all $n > N_\epsilon$ we have $U(f, P_n) - L(f, P_n) < \epsilon$. We know that $U(f, P_n) \geq U(f)$ and $L(f, P_n) \leq L(f)$. Therefore, we have $U(f) - L(f) \leq U(f, P_n) - L(f, P_n) < \epsilon$, which implies that $U(f)$ and $L(f)$ get arbitrarily close, as desired.

This $P_n$ is not necessarily related to each other, and $U(f, P_n)$ need not be monotonic.

Practically speaking, we can often show integrability by proposing a sequence $P_n$ (or an arbitrary one) and showing that $\lim U(f, P_n) - L(f, P_n) = 0$ using classical epsilon approaches.

Improper Integrals 🚀 (bonus)

So what if you want to take $\int_a^b f(x)dx$ but $f(b) = \infty$? In this case, it is not Darboux integrable, but this doesn’t necessairly mean that the whole thing isn’t integrable. You can try to compute $F(t) = \int_a^t f(x)dx$. if you can show that $F(t)$ is finite for all $t \in [a, b)$, then you can take $\lim_{t → b}F(t)$. It can be that as you get closer, this $F$ starts to diverge. You’ll see this if you have some $\epsilon$ in the denominator. However, if you have all $\epsilon$ in the numerator, this $F$ may converge. In this case, then you do have a value of $\int_a^b f(x)dx$.

Implicitly, we assume that $F(t)$ is continuous at $t = b$. So, you need to make sure that this is true by looking at the formula for $F(t)$.

If you have an infinite discontinuity in the middle of an interval, just split the integral.

Continuous Integrals 🚀 (bonus)

You can show that as long as $f$ is bounded, the antiderivative is continuous. In other words, $\lim_{x_0}F(x) = F(x_0)$. You can do this by bounding $\int_a^x -M dx \leq \int_a^x f(x)dx \leq \int_a^x Mdx$, and this squeezes the middle integral to be $F(x) → F(a)$ (because it’s $F(x) - F(a)$ to start with.

Meshes and Integrability

Mesh Definition 🐧

We define a mesh of a partition to be $\max_k (x_k - x_{k-1})$, i.e. what is the coarsest gap you have in your partition?

Cauchy Integrability Criterion 🚀 ⭐

Theorem: a bounded function ($|f(x)| < B$) is integrable if and only if for every $\epsilon > 0$, there exists a $\delta > 0$ such that

• Proof (manipulation of the summations)

  Backward direction is trivial. Any $P$ with $mesh(P) < \delta$ satisfies the integrability constraint. The forward direction takes a lot more work.

  First, because $f$ is integrable, let’s take some $P_0$ where $U(f, P_0) - L(f, P_0)< \epsilon/2$. We have some value $mesh(P_0)$. Intuitively, we might think that all $P$ whose $mesh(P) < mesh(P_0)$ have a better approximation to the integral, but that is actually quite tricky. If an interval of $P$ is wholly contained (or equal to) by an interval of $P_0$, then it is true. But if we are choosing an arbitrary $P$, this is not immediately obvious. We need more work.

  We can split up the $U(f, P) - L(f, P)$ into two sum componetns: one with intervals contained by $P_0$, and another with intervals that cross intervals by $P_0$:

  $$\sum_{contains}[M(f, I_P) - m(f, I_P)]|I_P|+\sum_{not contains}[M(f, I_P) - m(f, I_P)]|I_P|$$

  The first one we can replace with the interval of $P_0$ that contains the mentioned interval. Because this interval is larger, it is necessairly true that the replaced interval must have at least the range as the original interval. Furthermore, because we only included a part of $P_0$ in the summation, it must be $< \epsilon /2$. Let’s write this out:

  $$\sum_{contains}[M(f, I_P) - m(f, I_P)]|I_P| \leq \sum_{contains}[M(f, I_{P_0}) - m(f, I_{P_0})]|I_p| \leq U(f, P_0) - L(f, P_0) < \epsilon /2$$

  So, we’ve established that the first summation is less than $\epsilon / 2$. What about the second one?

  Well, recall that we haven’t used the $\delta$. In our search for the $P$, let’s define $\delta = \epsilon / 4Bn$, where $n$ is the number of partitions in $P_0$. First, we note that the number of elements in $\sum_{not contains}$ must be less than $n$, because you can only cross a partition boundary once. Second, we note that $M(f, I’) - m(f, I’) \leq 2 B$, because the function is bounded. Therefore, we can assemble the summation as follows:

  $$\sum_{not contains}[M(f, I_P) - m(f, I_P)]|I_P| \leq 2B \sum \frac{\epsilon}{4Bn} \leq 2B\frac{\epsilon}{4Bn}n = \epsilon / 2$$

  And therefore, the whole thing is $U(f, P) - L(f, P) < \epsilon$, as desired.

  Just a recap: we started with a valid partition $P_0$, and we showed that with a fine enough mesh, any $P$ can satisfy the epsilon constraint. We derive this using a helper partition, $P_0$.

The cauchy integrability criterion allows us to judge integrability in another way. We previously had to define a series of $P_n$. Now, we can just derive a generic mesh. Often, the most convenient mesh is the uniform one: $x_k = a + \frac{b - a}{n}k$. So, the cauchy integrability criterion enables you to use this uniform mesh to decide integrabiilty.

This is also why Darboux and Riemann integrals are equivalent.

Theories of Integrals

Integral Basic Theorems 🚀

Theorem:

• Proof sketch (definitions)

  You can write out the Darboux limit $\lim U(f + g, P_n)$ for any convergent $P_n$, and you can use the property of supremum that $\sup(a + b) \leq \sup(a) + \sup(b)$, and $\inf(a + b) \geq \inf(a) + \inf(b)$. You can also use properties of limit sums to help you out. But essentially you should be able to get that $U(f + g, P) - L(f + g, P) < \epsilon$, which shows integrability. Using the inequality $U(f + g) \leq U(f + g, P) \leq U(f, P) + U(g, P) < L(f, P) + L(g, P) + \epsilon \leq L(f) + L(g) + \epsilon$, we can get the original expression.

Theorem:

• Proof sketch (definitions)

  You can just use the fact that $\inf(c* x) = c \inf(x)$ and so forth. For negatives, we need to recognize that $\inf(-f) = \sup(f)$. And then you just reason by flipping the $L$ and the $U$.

Theorem:

• Proof sketch (definitions)

  Just weld together the partitions and see where that gets you.

Bounded Continuous Integrability Theorem 🚀

Theorem: if a function is bounded and continuous on $[a, b]$, then it is integrable on $[a, b]$.

• Proof (formal definition of integrability)

  Because $f$ is continuous on $[a, b]$, then it is uniformly continuous. So, for every $\epsilon$, there exists some $\delta_\epsilon$ such that if $|x - y| < \delta_\epsilon$, then we have $|f(x) - f(y)| < \epsilon$. Do you see where we are going here? Let’s make each partition $[x_{k-1}, x_k]$ such that $x_k - x_{k-1} < \delta_\epsilon$. Then, we have

  $$U(f, P_\epsilon) - L(f, P_\epsilon) = \sum (M(f[x_{k-1}, x_k]) - m(f[x_{k-1}, x_k]))(x_k - x_{k-1})$$

  Now, because of the uniform continuity statement, we know that $M(f[x_{k-1}, x_k]) - m(f[x_{k-1}, x_k]) < \epsilon$. So…we have

  $$U(f, P_\epsilon - L(f, P_\epsilon) \leq \epsilon \sum(x_k - x_{k-1}) = \epsilon (b - a)$$

  So, we can just set $\epsilon’ = \epsilon / (b - a)$, and we would have shown that there exists a partition $P_\epsilon$ for every $\epsilon$. This shows integrability.

  You can also just provide a concrete partition sequence, where $x_k = a + k\frac{b - a}{n}$, and in this case, you can actually show the existsnce of an $N_\epsilon$, etc. But more importantly is the above derivation with the $\epsilon$.

Bounded Monotone Integrability Theorem 🚀

While all bounded continuous functions are integrable, there are more functions that are integrable that are not continuous.

Theorem: all bounded monotone functions are integrable.

• Proof (formal definitions of integrability + monotone)

  Similar setup. Now, we can’t do anything fun with the functions just yet, but we can write out the difference: Without loss of generality, let’s assume that $f$ is increasing and that $|f(x)| < M$.

  $$U(f, P_\epsilon) - L(f, P_\epsilon) = \sum (M(f[x_{k-1}, x_k]) - m(f[x_{k-1}, x_k]))(x_k - x_{k-1})$$

  However, because $f$ is monotonic, we know that $M(f[x_{k-1}, x_k]) = f(x_{k-1})$. Similarly we know that $m(f[x_{k-1}, x_k]) = f(x_k)$. Therefore, the summation becomes

  $$U(f, P_\epsilon) - L(f, P_\epsilon) = \sum (f(x_k) - f(x_{k-1}))(x_k - x_{k-1})$$

  And if we assume that $x_k - x_{k-1}$ is a uniform distance, let’s say $\frac{1}{n}$, then we have

  $$U(f, P_\epsilon) - L(f, P_\epsilon) = \frac{1}{n}\sum (f(x_k) - f(x_{k-1}))$$

  and this telescopes to become

  $$U(f, P_\epsilon) - L(f, P_\epsilon) = \frac{1}{n}f(b) - f(a) \leq \frac{M}{n}$$

  So, if we let $n > \frac{M}{\epsilon}$, we will have constructed our partition.

  Concretely, this means that $P_\epsilon$ has uniform gaps with $x_k = a + k\frac{b - a}{n}$, where $n > M /\epsilon$.

  Because such a $P_\epsilon$ exists, we can conclude that $f$ is integrable.

Point Difference Integrability Theorem 🚀

Theorem: if $f, g$ are bounded functions and $f(x) = g(x)$ on $[a, b]$ except for a finite number of points, then if $f$ is integrable, then $g$ is integrable.

This is not true if the number of points is infinite. The reason is that, in the proof below, we rely on a finite partition and we can’t do this for an infinite number of points.

• Proof (definitions)

  We use a simple proof by induction, although we won’t actually do the inductive step. Let’s just do the base case, where we have one point different. The induction follows, although it is symbolically heavy.

  Assume that $|f| < B, |g| < B$.

  Because $f$ is integrable there exists some $P$ such that $U(f, P) - L(f, P) < \epsilon / 2$. Now, let $c$ be the place where $f(c) \neq g(c)$. Let’s consider a subpartition where we just add anothe partition $c \in[x_{\alpha}, x_\beta]$ around this point, such that $x_\beta - x_{\alpha} < \epsilon / 4B$. Let $Q$ be this new partition.

  From Darboux theorems, we know that $U(f, Q) - L(f, Q) \leq U(f, P) - L(f, P) < \epsilon / 2$.

  Ok, so we can express the following:

  $$U(g, Q) - L(g, Q) = \sum_{\text{exclude c}} [M(g, [x_{k-1}, x_k]) - m(g, [x_{k-1}, x_k])](x_k - x_{k-1}) + [M(g, [x_{\alpha}, x_\beta]) - m(g, [x_{\alpha}, x_\beta])](x_\beta - x_{\alpha})$$

  Now, we know that the first $\sum$ is the same as that of $f$, because we know that $g = f$ when it’s not $c$. Furthermore, we know that, because we exclude one element, we know that $\sum_{\text{exclude c}} [M(g, [x_{k-1}, x_k]) - m(g, [x_{k-1}, x_k])](x_k - x_{k-1}) < \epsilon / 2$.

  Next, we realize that because $g$ is bounded, we have

  $$[M(g, [x_{\alpha}, x_\beta]) - m(g, [x_{\alpha}, x_\beta])](x_\beta - x_{\alpha}) \leq (B -(-B))\epsilon / 4B = \epsilon / 2$$

  Therefore, if you add the two terms together, we get that $U(g, Q) - L(g, Q) < \epsilon$.

  Now, by the Darboux definition, because we can find such a partition $Q$ for every $\epsilon$ through this algorithm, $g$ is integrable.

  The induction follows easily. Just add another point!

Mean Value Theorem of Integrals 🚀

Theorem: if $f$ is continuous on $[a, b]$, then there must exist one $x \in (a, b)$ such that

Intuitively this is true if you think about the right hand side as the average value of a function. Basically, this is saying that $f$ must pass through the average value.

• Proof (Darboux conditions)

  We know the following is true:

  $$\int_a^bf(x)dx = U(f) \leq U(f, P)$$

  So, let’s pick the partition $[a, b]$, the coarsest possible. This means that $\int_a^b f(x)dx \leq M(f, [a, b])(b - a)$. We can do the same thing with the $L$ to get $\int_a^b f(x)dx \geq m(f, [a, b])(b - a)$.

  Therefore, we get

  $$m(f, [a, b]) \leq \frac{1}{b - a}\int_a^bf(x)dx \leq M(f, [a, b])$$

  And by the intermediate value theorem, we know that there must exist some $x_0$ such that $f(x_0) = \frac{1}{b-a}\int_a^b f(x)dx$.

Limits and integrals (bonus)

In some cases, you can switch limits and integrals. However, this may not be true all the time.

Fundamental Theorem of Calculus

The Fundamental Theorem 🚀

Theorem: if $f$ is continuous on $[a, b]$, let $g(x) = \int_a^x f(x)dx$. Then, $g’(x) = f(x)$. In other words, the derivative is the left inverse of the integral

• Proof (derivative definitions + mean value theorem)

  We know that $g’(x) = \lim_{h→0} \frac{1}{h} [g(x+h) - g(h)]$. By integral laws, we get $g’(x) = \lim_{h→0}\frac{1}{h} \int_x^{x+h}f(x)dx$.

  Now, by the mean value theorem, we know that there exists some $z(h)$ such that $f(z(h)) = \frac{1}{x + h - x} \int_x^{x+h}f(x)dx$. Therefore, the limit becomes

  $$g'(x) = \lim_{h\rightarrow0} f(z(h))$$

  Now, because $f$ is continuous, we can use the epsilon definition to finish off this proof. Pick an epsilon. There exists an $h$ such that if $\tilde x \in (x, x + h)$, then $|f(\tilde x) - f(x)| < \epsilon$. By the mean value theorem, we know that $z(h) \in (x, x + h)$, which means that $|f(z(h)) - f(x)| < \epsilon$. By definition, this means that $\lim_{h→0}f(z(h)) = f(x)$, as desired.

Theorem: if $f$ is continuous on $[a, b]$, then $\int_a^b f’(x)dx = f(b) - f(a)$. In other words, the derivative is also the right inverse of the integral.

• Proof sketch only

  Essentlaly, you can use the partitions to bound the integral with an $\epsilon$ distance of $g(b) - g(a)$, using the mean value theorem to translate between derivative and functin.

Integration by Parts 🚀

Theorem:

Why is this useful? Well, if we ever see a problem that looks like the product of two things, we can just write it as

• Proof (definitions)

  Let $g = uv$. We know that $g’ = u’v + uv’$. By the fundametnal theorem of calculus, we have

  $$\int_a^b g'(x)dx= \int_a^b u(x)v'(x) + u'(x)v(x)dx = g(b) - g(a) = u(a)v(a) - u(b)v(b)$$

U-substitution 🚀

Theorem:

Why do we care? Well, suppose that we had a funtion that looks like $f(u(x)) u’(x)$. We can change it into a simpler integral by essentially reversing the chain rule.

• Proof (chain rule)

  Let $F(u) = \int_c^uf(t)dt$. By the fundamental theorem, we have $F’(u) = f(u)$. Now, let $g = F\circ u$.

  By the chain rule, we have $g’(x) = F’(u(x)) u’(x) = f(u(x))u’(x)$.

  Therefore, we have by the fundamental rule again, that

  $$\int_a^b f(u(x))u'(x) = \int_a^bg'(x) = g(b) - g(a) = F(u(b)) - F(u(a)) = \int_{u(b)}^{u(a)}f(t)dt$$