On Measures, a positive measure was a countably additive function from $\mathcal{M}$ to $[0, \infty]$. Two things were baked in: the values are never negative, and $\infty$ is allowed.
So far every measure on this site has been positive. Mass, length, probability: they only pile up. But plenty of things you'd like to measure come with a sign, or a direction. Charge on a wire is positive in some places and negative in others. The integral $E \mapsto \int_E f \, dm$ of a function that changes sign goes up and down. And once the function is complex-valued, its integrals point in every direction of the plane.
This page builds the measures that allow that, the complex measures, and then answers the question they raise right away: if the pieces can cancel each other, how much is really there? The answer is the total variation $|\mu|$. It's a positive measure, it's always finite, and it turns the complex measures on $\mathcal{M}$ into a normed space. For real measures it splits $\mu$ cleanly into a positive part and a negative part.
Throughout, $(X, \mathcal{M})$ is a measurable space. Countable additivity talks about cutting a set into countably many measurable pieces, so give that a name first.
A measurable partition of $E \in \mathcal{M}$ is a sequence $(E_n)_{n=1}^\infty$ in $\mathcal{M}$ such that $E_i \cap E_j = \emptyset$ for all $i \ne j$ and $\bigcup_{n=1}^\infty E_n = E$.
Empty pieces are allowed, so a finite partition $E_1, \ldots, E_N$ counts too: pad it with $\emptyset$. Now keep the definition of a measure and just change where the values live.
A complex measure on $\mathcal{M}$ is a function $\mu\colon \mathcal{M} \to \mathbb{C}$ such that $$\mu(E) = \sum_{n=1}^\infty \mu(E_n)$$ for every $E \in \mathcal{M}$ and every measurable partition $(E_n)_{n=1}^\infty$ of $E$.
Read the codomain carefully: $\mathbb{C}$. A complex measure never takes the value $\infty$, and the series on the right is a series of complex numbers that has to converge. For a positive measure, a divergent sum was fine: it just meant $\infty$. Here there's nowhere for it to go.
A positive measure isn't a complex measure. It's allowed to be infinite and complex measures aren't. Rich-person problems.
So Lebesgue measure on $\mathbb{R}$ is not a complex measure, but Lebesgue measure on $[0, 1]$ is. There's a second, sneakier consequence. A partition has no built-in order: list its pieces in any order you like and it's the same partition of the same set. So the series has to give the same answer in every order. That's a strong demand.
The alternating harmonic series converges, but only conditionally, and rearranging it changes its sum. A complex measure can't afford that, and it doesn't have to worry about it either.
If $\mu$ is a complex measure and $(E_n)$ is a measurable partition of $E$, then $\sum_{n=1}^\infty |\mu(E_n)| < \infty$.
Proof. Every reordering of a measurable partition $(E_n)$ of $E$ is again a measurable partition of $E$, so $\sum_n \operatorname{Re} \mu(E_n)$ converges to $\operatorname{Re} \mu(E)$ in every order. By Riemann's rearrangement theorem, a real series that converges in every order converges absolutely, so $\sum_n |\operatorname{Re} \mu(E_n)| < \infty$. The same argument gives $\sum_n |\operatorname{Im} \mu(E_n)| < \infty$, and $|\mu(E_n)| \le |\operatorname{Re} \mu(E_n)| + |\operatorname{Im} \mu(E_n)|$. $\blacksquare$
Absolute convergence will matter all over this page: it's what lets us rearrange and regroup the pieces freely. When the values happen to be real, the measure gets its own name.
A real measure is a complex measure whose values are all real numbers.
A real measure can be negative, but it can't be infinite. That's the difference from a positive measure, which can be infinite but can't be negative. A finite positive measure is both.
Other books call a real measure a "signed measure." Watch the fine print when you read them: a signed measure is often allowed to take one of the values +∞ or −∞ (never both), so not every signed measure is a real measure in the sense used here.
Here's the problem with signed and complex values: the number $\mu(E)$ can badly undersell how much is going on inside $E$. Positive mass in one part of $E$ cancels negative mass in another. To see all of it, cut $E$ into pieces, measure each piece, and add up the sizes. The finer you cut, the less can cancel inside a piece. So take the best you can do.
Let $\mu$ be a complex measure on $\mathcal{M}$. The total variation of $\mu$ is the function $|\mu|\colon \mathcal{M} \to [0, \infty]$ given by $$|\mu|(E) = \sup\left\{\sum_{n=1}^\infty |\mu(E_n)| : (E_n)_{n=1}^\infty \text{ is a measurable partition of } E\right\}.$$
The partition $\{E\}$ (with empty pieces after it) is one of the partitions in the sup, so $|\mu(E)| \le |\mu|(E)$ always. The gap between the two is exactly the cancelling. Here's the simplest case where they disagree.
On the Borel sets of $\mathbb{R}$, let $\mu = \delta_0 - \delta_1$, so $\mu(E) = \delta_0(E) - \delta_1(E)$. Find $\mu(E)$ and $|\mu|(E)$ for $E = \{0, 1\}$.
$\mu(E) = 1 - 1 = 0$.
For $|\mu|(E)$: the partition $\{0\}, \{1\}$ gives $|\mu(\{0\})| + |\mu(\{1\})| = 1 + 1 = 2$. No partition does better: in any partition, one piece contains $0$ and one piece contains $1$. Every other piece has measure $0$, and the pieces holding $0$ and $1$ have measure $1$, $-1$, or (if they're the same piece) $0$. So the sum is at most $2$, and $|\mu|(E) = 2$.
The measure of $E$ is $0$, but there's plenty there: a unit of mass at $0$ and a unit of anti-mass at $1$. The bars sit around the $\mu$, not around $\mu(E)$, and that's the whole point.
$|\mu|(E)$ and $|\mu(E)|$ look almost the same on the page and mean different things. $|\mu(E)|$ is the size of one number, after all the cancelling. $|\mu|(E)$ is the total variation measure evaluated at $E$, before any cancelling. You only get $|\mu|(E) = |\mu(E)|$ when nothing in $E$ cancels.
A sup over every partition sounds hard to compute. For the measures we'll meet most, it isn't. The first kind is a measure made of point masses: write $\mu = \sum_{x \in D} w_x \delta_x$ for the measure that puts the complex weight $w_x$ at each point $x$ of a countable set $D$.
Let $D \subseteq X$ be countable with $\{x\} \in \mathcal{M}$ for every $x \in D$, and let $(w_x)_{x \in D}$ be complex numbers with $\sum_{x \in D} |w_x| < \infty$. Then $\mu(E) = \sum_{x \in E \cap D} w_x$ is a complex measure, and $$|\mu|(E) = \sum_{x \in E \cap D} |w_x| \quad \text{for every } E \in \mathcal{M}.$$
Proof. Each $\mu(E)$ is a subseries of an absolutely convergent series, so it's a well-defined complex number. If $(E_n)$ is a measurable partition of $E$, each $x \in E \cap D$ lies in exactly one $E_n$, so $\sum_n \mu(E_n) = \sum_n \sum_{x \in E_n \cap D} w_x = \sum_{x \in E \cap D} w_x = \mu(E)$: regrouping an absolutely convergent series doesn't change its sum. So $\mu$ is a complex measure.
Let $(E_n)$ be a measurable partition of $E$. Then $|\mu(E_n)| = \big|\sum_{x \in E_n \cap D} w_x\big| \le \sum_{x \in E_n \cap D} |w_x|$, and each $x \in E \cap D$ lies in exactly one $E_n$, so $\sum_n |\mu(E_n)| \le \sum_{x \in E \cap D} |w_x|$. Taking the sup, $|\mu|(E) \le \sum_{x \in E \cap D} |w_x|$.
For the other inequality, $D$ is a countable union of measurable singletons, so $E \setminus D \in \mathcal{M}$. The singletons $\{x\}$ with $x \in E \cap D$, together with $E \setminus D$ (and empty sets, if $E \cap D$ is finite), form a measurable partition of $E$. Since $\mu(E \setminus D) = 0$, its sum is $\sum_{x \in E \cap D} |w_x|$. $\blacksquare$
So atoms count with their moduli: put every atom in its own piece and nothing cancels. Here it is with numbers.
On the Borel sets of $\mathbb{R}$, let $\mu = 3\delta_0 + 4i\,\delta_1 - 2\delta_5$. Find $\mu([0, 1])$, $|\mu([0, 1])|$, $|\mu|([0, 1])$, and $\|\mu\|$.
The atoms in $[0, 1]$ are at $0$ and $1$, so $\mu([0, 1]) = 3 + 4i$ and $|\mu([0, 1])| = \sqrt{3^2 + 4^2} = 5$.
By the proposition, $|\mu|([0, 1]) = |3| + |4i| = 7$: the two weights point in different directions, so lumping them together loses $2$.
$\|\mu\| = |\mu|(\mathbb{R}) = 3 + 4 + 2 = 9$, while $|\mu(\mathbb{R})| = |1 + 4i| = \sqrt{17}$.
Try to beat the proposition by grouping, and watch it fail.
Grouping only hurts when the arrows in a group point different ways. Weights pointing the same way can share a piece for free. That's the whole story for atoms. The second kind of measure has a density.
On The Lebesgue Integral, a measurable $f \ge 0$ gave a positive measure $E \mapsto \int_E f \, d\lambda$, written $d\nu = f \, d\lambda$. Now $f$ is allowed to be negative, as long as it's integrable.
Let $\lambda$ be a positive measure on $\mathcal{M}$ and $f \in L^1(\lambda)$ real-valued. Then $\mu(E) = \int_E f \, d\lambda$ is a real measure, written $d\mu = f \, d\lambda$, and $$|\mu|(E) = \int_E |f| \, d\lambda \quad \text{for every } E \in \mathcal{M}.$$
Proof. $\mu(E) = \int_E f^+ \, d\lambda - \int_E f^- \, d\lambda$. Each term is a positive measure in $E$ (by the density theorem on The Lebesgue Integral), and each is finite because $f \in L^1(\lambda)$. A difference of two finite, countably additive functions is countably additive, so $\mu$ is a real measure.
For a measurable partition $(E_n)$ of $E$, $|\mu(E_n)| = \big|\int_{E_n} f \, d\lambda\big| \le \int_{E_n} |f| \, d\lambda$. Since $E \mapsto \int_E |f| \, d\lambda$ is a positive measure, $\sum_n \int_{E_n} |f| \, d\lambda = \int_E |f| \, d\lambda$. So every partition sum is at most $\int_E |f| \, d\lambda$, and so is $|\mu|(E)$.
For the other inequality, let $P = \{f > 0\}$ and partition $E$ into $E \cap P$ and $E \setminus P$. Then $|\mu(E \cap P)| + |\mu(E \setminus P)| = \int_{E \cap P} f \, d\lambda - \int_{E \setminus P} f \, d\lambda = \int_E |f| \, d\lambda$, so $|\mu|(E) \ge \int_E |f| \, d\lambda$. $\blacksquare$
Same moral: cut where the sign changes and nothing cancels. Here's what happens when you don't know where the sign changes and just keep cutting. The density is $f(x) = 2\sin(3\pi x)$ on $[0, 1]$: two positive humps and a negative one, so $|\mu|([0, 1]) = \int_0^1 |f| \, dm = \frac{4}{\pi}$.
Cutting a piece into smaller pieces can only help, by the triangle inequality. But "more pieces" isn't the same as "finer": $4$ equal pieces don't refine $3$ equal pieces. Find a case where adding a piece makes things worse.
Refining helps, and that's the key step in showing that $|\mu|$ is a measure. Here it is on its own.
Let $\mu$ be a complex measure and let $(A_m)$ and $(E_n)$ be measurable partitions of the same set $E$. Prove that $$\sum_m |\mu(A_m)| \le \sum_n |\mu|(E_n).$$
Proof. For each $m$, the sets $A_m \cap E_n$ ($n = 1, 2, \ldots$) form a measurable partition of $A_m$, so $\mu(A_m) = \sum_n \mu(A_m \cap E_n)$. By the triangle inequality for series, and then swapping the order of a double series of nonnegative terms, $$\sum_m |\mu(A_m)| = \sum_m \Big|\sum_n \mu(A_m \cap E_n)\Big| \le \sum_n \sum_m |\mu(A_m \cap E_n)| \le \sum_n |\mu|(E_n),$$ because for each $n$ the sets $A_m \cap E_n$ ($m = 1, 2, \ldots$) form a measurable partition of $E_n$, so their sum is one of the sums in the sup that defines $|\mu|(E_n)$. $\blacksquare$
Now the first big fact. Total variation was defined by a sup, with no promise that it adds up. It does.
If $\mu$ is a complex measure on $\mathcal{M}$, then $|\mu|$ is a positive measure on $\mathcal{M}$.
Proof. The values are in $[0, \infty]$, and $|\mu|(\emptyset) = 0$ because every piece of a partition of $\emptyset$ is empty. Let $(E_n)$ be a measurable partition of $E$.
($\ge$) For each $n$, let $t_n$ be a real number with $0 \le t_n < |\mu|(E_n)$ (or $t_n = 0$ if $|\mu|(E_n) = 0$), and choose a measurable partition $(A_{n,m})_m$ of $E_n$ with $\sum_m |\mu(A_{n,m})| \ge t_n$. Then $$\sum_n t_n \le \sum_n \sum_m |\mu(A_{n,m})| \le |\mu|(E),$$ because the sets $A_{n,m}$, over all $n$ and $m$, form a measurable partition of $E$ (list them as one sequence: a sum of nonnegative terms doesn't depend on the order). Taking the sup over all such choices of the $t_n$ gives $\sum_n |\mu|(E_n) \le |\mu|(E)$.
($\le$) For every measurable partition $(A_m)$ of $E$, the Exercise above gives $\sum_m |\mu(A_m)| \le \sum_n |\mu|(E_n)$. Taking the sup over $(A_m)$ gives $|\mu|(E) \le \sum_n |\mu|(E_n)$. $\blacksquare$
So $|\mu|$ is a genuine measure, sitting on top of $\mu$ and dominating it: $|\mu(E)| \le |\mu|(E)$ for every $E$. One sanity check before moving on: if $\mu$ was positive to begin with, nothing can cancel.
Let $\mu$ be a positive measure with $\mu(X) < \infty$. Show that $\mu$ is a complex measure and that $|\mu| = \mu$.
Proof. Every value satisfies $0 \le \mu(E) \le \mu(X) < \infty$, so $\mu$ maps into $\mathbb{C}$, and it's countably additive with convergent series, so it's a complex measure. For any measurable partition $(E_n)$ of $E$, $$\sum_n |\mu(E_n)| = \sum_n \mu(E_n) = \mu(E).$$ Every partition gives the same sum $\mu(E)$, so the sup is $\mu(E)$: $|\mu|(E) = \mu(E)$. $\blacksquare$
Time to compute some. One piece of shorthand: $\mu = f \, dm + \sum_x w_x \delta_x$ means the measure $E \mapsto \int_E f \, dm + \sum_{x \in E} w_x$. On a set the atoms live on, the density contributes nothing (a countable set has Lebesgue measure $0$), so the two propositions combine: $|\mu|(E) = \int_E |f| \, dm + \sum_{x \in E} |w_x|$.
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We still owe one fact: is $|\mu|(X)$ finite? A positive measure can be infinite, and $|\mu|$ is a positive measure, so this is a real question. The answer is yes, and the proof runs through a lemma about finitely many complex numbers. It says you can always pick some of them so that their sum keeps a fixed fraction of their total length. Cancelling can't eat everything.
If $z_1, \ldots, z_N \in \mathbb{C}$, then there is a subset $S \subseteq \{1, \ldots, N\}$ such that $$\Big|\sum_{k \in S} z_k\Big| \ge \frac{1}{\pi} \sum_{k=1}^N |z_k|.$$
Proof. Write $z_k = |z_k| e^{i\alpha_k}$, and let $\cos^+ x = \max\{\cos x, 0\}$. For $\theta \in [0, 2\pi]$, let $S(\theta) = \{k : \cos(\alpha_k - \theta) > 0\}$. Then $$\Big|\sum_{k \in S(\theta)} z_k\Big| = \Big|\sum_{k \in S(\theta)} e^{-i\theta} z_k\Big| \ge \operatorname{Re} \sum_{k \in S(\theta)} e^{-i\theta} z_k = \sum_{k \in S(\theta)} |z_k| \cos(\alpha_k - \theta) = \sum_{k=1}^N |z_k| \cos^+(\alpha_k - \theta).$$ Average both sides over $\theta \in [0, 2\pi]$. By periodicity, for each $k$, $$\frac{1}{2\pi} \int_0^{2\pi} \cos^+(\alpha_k - \theta) \, d\theta = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} \cos t \, dt = \frac{1}{\pi},$$ so the average of the left side is at least $\frac{1}{\pi} \sum_k |z_k|$. The left side, as a function of $\theta$, takes only finitely many values (one for each subset), so it has a maximum, and a maximum is at least the average. At a $\theta_0$ where it's largest, $S = S(\theta_0)$ works. $\blacksquare$
The idea: point a direction $e^{i\theta}$ and keep every $z_k$ that makes an acute angle with it. Those can't cancel each other much, because they all lean the same way. Watch the direction go once around.
The graph on the right dips under the green level, so not every direction works. The proof never finds the best direction. It averages over all of them instead: each $z_k$ contributes a hump $|z_k| \cos^+(\alpha_k - \theta)$, and the humps are easy to average.
$\pi$ showed up in a lemma about adding a finite list of numbers. Nobody invited $\pi$. $\pi$ came anyway.
It came from averaging a cosine over a full turn. Now try it yourself. Drag the numbers around and turn the half-plane; see how low you can push the best ratio.
The lemma only promises some subset; it picks a half-plane's worth because that's easy to analyze. Is that a loss? Pick subsets by hand and compare with the best one.
The best subset always turns out to be the arrows on one side of a line. And spreading the arrows evenly drags the best ratio down toward $\frac{1}{\pi}$. That's no accident: $\frac{1}{\pi}$ is the best constant there is.
Let $N$ be even and $z_k = e^{2\pi i k/N}$ for $k = 1, \ldots, N$, so $\sum_k |z_k| = N$.
Proof. (a) With $\omega = e^{2\pi i/N}$, $m$ consecutive terms add to $z_j(1 + \omega + \cdots + \omega^{m-1}) = z_j \frac{1 - \omega^m}{1 - \omega}$, and $|1 - e^{i\phi}| = 2|\sin(\phi/2)|$, so the modulus is $\frac{\sin(\pi m/N)}{\sin(\pi/N)}$ for $0 \le m \le N$, which is at most $\frac{1}{\sin(\pi/N)}$. The condition $\cos(\alpha_k - \theta) > 0$ says the angle of $z_k$ lies in the open arc $(\theta - \frac{\pi}{2}, \theta + \frac{\pi}{2})$ of length $\pi$. The $z_k$ are spaced $\frac{2\pi}{N}$ apart, so that arc holds at most $N/2$ of them, all consecutive.
(b) Let $S$ be any subset with $s = \sum_{k \in S} z_k \ne 0$, and let $\theta$ be the angle of $s$. Then $$|s| = \operatorname{Re}(e^{-i\theta} s) = \sum_{k \in S} \operatorname{Re}(e^{-i\theta} z_k) \le \sum_{k \in S(\theta)} \operatorname{Re}(e^{-i\theta} z_k) \le \Big|\sum_{k \in S(\theta)} z_k\Big| \le \frac{1}{\sin(\pi/N)},$$ since passing from $S$ to $S(\theta)$ drops the terms with $\operatorname{Re}(e^{-i\theta} z_k) \le 0$ and adds the missing positive ones, and then (a) applies. $N/2$ consecutive roots reach $\frac{\sin(\pi/2)}{\sin(\pi/N)} = \frac{1}{\sin(\pi/N)}$, so that's the max. The best ratio is $\frac{1}{N \sin(\pi/N)}$, and $N \sin(\pi/N) \to \pi$ as $N \to \infty$. So for any $c > \frac{1}{\pi}$, a large even $N$ has every subset below $c \sum_k |z_k|$. $\blacksquare$
Practice the pick from the proof: given the numbers and a direction, find the sum the half-plane keeps.
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Now cash in the lemma. If $|\mu|(X)$ were infinite, there would be huge partition sums, and the lemma would let us carve off a piece with a big value of $\mu$. Do that over and over and countable additivity breaks.
If $\mu$ is a complex measure on $\mathcal{M}$, then $|\mu|(X) < \infty$.
Proof. Claim: if $E \in \mathcal{M}$ has $|\mu|(E) = \infty$, then $E$ is the disjoint union of measurable sets $A$ and $B$ with $|\mu(A)| > 1$, $|\mu(B)| > 1$, and $|\mu|(A) = \infty$ or $|\mu|(B) = \infty$.
Let $t = \pi(1 + |\mu(E)|)$. Since $|\mu|(E) > t$, some measurable partition $(E_n)$ of $E$ has $\sum_n |\mu(E_n)| > t$, and then already $\sum_{n=1}^N |\mu(E_n)| > t$ for some $N$. The lemma, applied to $\mu(E_1), \ldots, \mu(E_N)$, gives $S \subseteq \{1, \ldots, N\}$ with $\big|\sum_{n \in S} \mu(E_n)\big| \ge \frac{1}{\pi} \sum_{n=1}^N |\mu(E_n)| > \frac{t}{\pi}$. Let $A = \bigcup_{n \in S} E_n$, so $|\mu(A)| > \frac{t}{\pi} = 1 + |\mu(E)| \ge 1$, and let $B = E \setminus A$. Then $$|\mu(B)| = |\mu(E) - \mu(A)| \ge |\mu(A)| - |\mu(E)| > \frac{t}{\pi} - |\mu(E)| = 1.$$ Finally $|\mu|$ is a measure, so $|\mu|(A) + |\mu|(B) = |\mu|(E) = \infty$, and one of the two is $\infty$.
The proposition. Suppose $|\mu|(X) = \infty$. The claim splits $X$ into two pieces; call the one with infinite total variation $B_1$ and the other $A_1$, so $|\mu(A_1)| > 1$ and $|\mu|(B_1) = \infty$. Apply the claim to $B_1$ to get $A_2$ and $B_2$ the same way, and keep going. The sets $A_1, A_2, \ldots$ are pairwise disjoint ($A_{i+1} \subseteq B_i$, which misses $A_1, \ldots, A_i$) and $|\mu(A_i)| > 1$ for every $i$. Countable additivity says $\mu\big(\bigcup_i A_i\big) = \sum_i \mu(A_i)$, but the terms $\mu(A_i)$ do not tend to $0$, so the series diverges. That's a contradiction, so $|\mu|(X) < \infty$. $\blacksquare$
Here's the argument as a picture. Each round splits off one more piece with $|\mu(A_i)| > 1$, and the infinite variation always survives in what's left.
So a complex measure is bounded: $|\mu(E)| \le |\mu|(E) \le |\mu|(X) < \infty$, and every value lies in one disk of finite radius. Some books say $\mu$ has bounded variation. Here's a check on all of it: which set functions are complex measures, and which only look like it?
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Complex measures can be added and scaled, and the result is still a complex measure, because convergent series add and scale. Bounded means we can also measure how big one is, so they form a normed space.
$\mathbb{M}(\mathcal{M})$ is the set of all complex measures on $\mathcal{M}$, with $(\mu + \alpha\nu)(E) = \mu(E) + \alpha\nu(E)$ for $\mu, \nu \in \mathbb{M}(\mathcal{M})$ and $\alpha \in \mathbb{C}$, and with the total variation norm $\|\mu\| = |\mu|(X)$.
On Banach Spaces & Bounded Operators, a norm had to be positive except at $0$, pull out scalars as $|\alpha|$, and satisfy the triangle inequality.
The name "norm" is a promise. Here's the proof that it's kept.
$\mathbb{M}(\mathcal{M})$ is a complex vector space, and $\|\mu\| = |\mu|(X)$ is a norm on it.
Proof. If $\mu, \nu$ are complex measures and $(E_n)$ partitions $E$, then $\sum_n (\mu + \alpha\nu)(E_n) = \sum_n \mu(E_n) + \alpha \sum_n \nu(E_n) = (\mu + \alpha\nu)(E)$, so $\mu + \alpha\nu$ is a complex measure; the vector space rules hold setwise. By the boundedness proposition, $\|\mu\|$ is a finite number $\ge 0$.
Zero only at zero. If $|\mu|(X) = 0$, then for every measurable $E$, $|\mu(E)| \le |\mu|(E) \le |\mu|(X) = 0$, using that the measure $|\mu|$ is monotone. So $\mu = 0$.
Scalars. For every partition, $\sum_n |\alpha\mu(E_n)| = |\alpha| \sum_n |\mu(E_n)|$; taking sups, $|\alpha\mu|(X) = |\alpha| \, |\mu|(X)$.
Triangle inequality. For a measurable partition $(E_n)$ of $E$, $\sum_n |(\mu + \nu)(E_n)| \le \sum_n |\mu(E_n)| + \sum_n |\nu(E_n)| \le |\mu|(E) + |\nu|(E)$. Taking the sup over partitions, $|\mu + \nu|(E) \le |\mu|(E) + |\nu|(E)$, and $E = X$ gives $\|\mu + \nu\| \le \|\mu\| + \|\nu\|$. $\blacksquare$
For measures made of atoms the norm is just $\sum_x |w_x|$, so you can feel the triangle inequality point by point. Make it an equality, then make it collapse.
Equality needs $\mu$ and $\nu$ to point the same way wherever both are nonzero. With $\nu = -\mu$, everything cancels, and that's the zero vector of the space.
On Duality: (Lᵖ)* and C₀(X)*, a later page, the regular complex Borel measures $M(X)$, a subspace of this space with the same norm, turn out to be exactly the dual of $C_0(X)$. The total variation norm is what makes $\|\Lambda\| = |\mu|(X)$ there.
For a real measure, $|\mu|$ counts the positive mass and the negative mass together, while $\mu$ counts them against each other. Average the two and you can pull them apart.
Let $\mu$ be a real measure. Its positive variation and negative variation are $$\mu^+ = \frac{1}{2}(|\mu| + \mu) \quad \text{and} \quad \mu^- = \frac{1}{2}(|\mu| - \mu).$$
By design $\mu^+ - \mu^- = \mu$ and $\mu^+ + \mu^- = |\mu|$. What isn't obvious from the formulas is that $\mu^+$ and $\mu^-$ are never negative.
If $\mu$ is a real measure, then $\mu^+$ and $\mu^-$ are finite positive measures, and $$\mu = \mu^+ - \mu^-, \qquad |\mu| = \mu^+ + \mu^-.$$ Writing $\mu = \mu^+ - \mu^-$ is the Jordan decomposition of $\mu$.
Proof. $\mu^+$ is countably additive because $|\mu|$ and $\mu$ are (both series converge, by the boundedness proposition), and it's finite for the same reason. It is never negative because $|\mu(E)| \le |\mu|(E)$, so $|\mu|(E) + \mu(E) \ge 0$. The same goes for $\mu^-$, with $|\mu|(E) - \mu(E) \ge 0$. The two identities are algebra. $\blacksquare$
Picture it as peeling: the positive mass goes one way, the negative mass flips over and goes the other, and putting them back side by side gives the total.
$\mu^+$ collects all the good news, $\mu^-$ all the bad news, and $|\mu|$ reads both and just says it was a lot of news.
For a density, the peeling is exactly the split $f = f^+ - f^-$ you already know. One example by hand.
Let $d\mu = (2x - 1) \, dm$ on the Borel subsets of $[0, 1]$. Find $\mu([0, 1])$, $|\mu|([0, 1])$, $\mu^+([0, 1])$, and $\mu^-([0, 1])$.
$\mu([0, 1]) = \int_0^1 (2x - 1) \, dx = [x^2 - x]_0^1 = 0$.
By the density proposition, $|\mu|([0, 1]) = \int_0^1 |2x - 1| \, dx$. The graph of $|2x - 1|$ is two triangles with base $\frac{1}{2}$ and height $1$, so $|\mu|([0, 1]) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
$\mu^+([0, 1]) = \frac{1}{2}\big(\frac{1}{2} + 0\big) = \frac{1}{4}$ and $\mu^-([0, 1]) = \frac{1}{2}\big(\frac{1}{2} - 0\big) = \frac{1}{4}$. Check: $\int_0^1 (2x - 1)^+ \, dx$ is the right-hand triangle, $\frac{1}{4}$. In fact $d\mu^+ = f^+ \, dm$ and $d\mu^- = f^- \, dm$, because $\frac{1}{2}(|f| + f) = f^+$ and $\frac{1}{2}(|f| - f) = f^-$.
Build your own real measure and peel it. Watch $\mu(E)$ hit $0$ while $\mu^+(E)$ and $\mu^-(E)$ stay big.
A real measure can be written as a difference of two positive measures in lots of ways: add the same finite measure to both parts. The Jordan decomposition is the most economical one.
Let $\mu$ be a real measure, and suppose $\mu = \lambda_1 - \lambda_2$ for finite positive measures $\lambda_1, \lambda_2$. Prove that $\mu^+ \le \lambda_1$ and $\mu^- \le \lambda_2$ (as set functions).
Proof. For a measurable partition $(E_n)$ of $E$, $|\mu(E_n)| = |\lambda_1(E_n) - \lambda_2(E_n)| \le \lambda_1(E_n) + \lambda_2(E_n)$, so $\sum_n |\mu(E_n)| \le \lambda_1(E) + \lambda_2(E)$. Taking the sup, $|\mu|(E) \le \lambda_1(E) + \lambda_2(E)$. Then $$\mu^+(E) = \tfrac{1}{2}\big(|\mu|(E) + \mu(E)\big) \le \tfrac{1}{2}\big(\lambda_1(E) + \lambda_2(E) + \lambda_1(E) - \lambda_2(E)\big) = \lambda_1(E),$$ and in the same way $\mu^-(E) \le \frac{1}{2}\big(\lambda_1(E) + \lambda_2(E) - \lambda_1(E) + \lambda_2(E)\big) = \lambda_2(E)$. $\blacksquare$
On The Radon–Nikodym Theorem, the next page, the Hahn decomposition finds disjoint sets $A$ and $B$ with $\mu^+(E) = \mu(E \cap A)$ and $\mu^-(E) = -\mu(E \cap B)$: the peeling happens along a cut of $X$ itself. The same page proves $|\mu|(E) = \int_E |h| \, d\lambda$ for complex densities too.
Last lesson question: compute a piece of the Jordan decomposition.
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Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Next, the proofs. Each one is a line from a proof on this page with a gap in it. Fill the gap.
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Then the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types that need their answer choices will say so.
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Last, the Spicy Bonus. We showed $\mathbb{M}(\mathcal{M})$ is a normed space. It's better than that: it's complete.
Prove that $\mathbb{M}(\mathcal{M})$, with the total variation norm $\|\mu\| = |\mu|(X)$, is a Banach space.
Proof. Let $(\mu_n)$ be a Cauchy sequence in $\mathbb{M}(\mathcal{M})$.
A limit setwise. For every $E \in \mathcal{M}$, $|\mu_n(E) - \mu_m(E)| \le |\mu_n - \mu_m|(E) \le \|\mu_n - \mu_m\|$. So $(\mu_n(E))$ is Cauchy in $\mathbb{C}$; let $\mu(E)$ be its limit. Every $\mu(E)$ is a complex number, and $\mu(\emptyset) = 0$.
A uniform estimate. Let $\varepsilon > 0$ and choose $N$ with $\|\mu_n - \mu_m\| < \varepsilon$ for all $n, m \ge N$. Let $(E_k)$ be a measurable partition of some $E$. For each $K$, the sets $E_1, \ldots, E_K$, $E \setminus (E_1 \cup \cdots \cup E_K)$ form a measurable partition of $E$, so for $n, m \ge N$, $$\sum_{k=1}^K |\mu_n(E_k) - \mu_m(E_k)| \le |\mu_n - \mu_m|(E) < \varepsilon.$$ Let $m \to \infty$ (a finite sum), then $K \to \infty$: $$\sum_{k=1}^\infty |\mu_n(E_k) - \mu(E_k)| \le \varepsilon \quad \text{for every } n \ge N. \tag{$*$}$$
$\mu$ is a complex measure. Fix $n \ge N$. The series $\sum_k \mu_n(E_k)$ converges absolutely to $\mu_n(E)$, and by $(*)$ the terms $\mu(E_k)$ differ from $\mu_n(E_k)$ by an absolutely summable sequence, so $\sum_k \mu(E_k)$ converges, and $$\Big|\sum_k \mu(E_k) - \mu(E)\Big| \le \sum_k |\mu(E_k) - \mu_n(E_k)| + \Big|\sum_k \mu_n(E_k) - \mu_n(E)\Big| + |\mu_n(E) - \mu(E)| \le \varepsilon + 0 + |\mu_n(E) - \mu(E)|.$$ Let $n \to \infty$: the right side tends to $\varepsilon$. Since $\varepsilon$ was arbitrary, $\sum_k \mu(E_k) = \mu(E)$.
$\mu_n \to \mu$ in norm. Apply $(*)$ with $E = X$: for $n \ge N$, every measurable partition $(E_k)$ of $X$ has $\sum_k |(\mu_n - \mu)(E_k)| \le \varepsilon$. Taking the sup, $\|\mu_n - \mu\| \le \varepsilon$ for all $n \ge N$. So $\mu_n \to \mu$ in $\mathbb{M}(\mathcal{M})$, and the space is complete. $\blacksquare$