On The Lebesgue Integral, a function $f \ge 0$ turned into a measure: $\nu(E) = \int_E f \, d\mu$. Integrals make measures.
Every measure built from a density has one telltale habit: if $\mu(E) = 0$, then $\int_E f \, d\mu = 0$, so the new measure ignores every set $\mu$ ignores. This page asks the converse. Hand me a measure $\lambda$ that ignores every $\mu$-null set. Is it secretly $\int_E h \, d\mu$ for some $h$?
The answer is the Lebesgue–Radon–Nikodym theorem, and it says more than yes: every complex measure splits into a part that comes from a density and a part that lives on a $\mu$-null set, and both parts are unique. Then three consequences fall out almost for free: the ε–δ picture of absolute continuity, the polar decomposition $d\mu = h \, d|\mu|$, and the Hahn decomposition, which cuts the space itself into a positive half and a negative half.
Throughout, $(X, \mathcal{M})$ is a measurable space and $\mu$ is a positive measure on it. The other measure, $\lambda$, can be positive or complex. First, the habit every density has, turned into a definition.
$\lambda$ is absolutely continuous with respect to $\mu$, written $\lambda \ll \mu$, if $\lambda(E) = 0$ for every $E \in \mathcal{M}$ with $\mu(E) = 0$.
Read the symbol as "$\lambda$ is dominated by $\mu$": the big measure $\mu$ decides which sets are negligible, and $\lambda$ has to agree.
The direction is the whole definition. $\lambda \ll \mu$ checks the null sets of $\mu$, never the null sets of $\lambda$. With $\lambda = \chi_{[0, 1/2]} \, dm$ and $\mu = m$ on $[0, 1]$, $\lambda \ll \mu$ is true, but $\mu \ll \lambda$ is false: $\lambda((\frac{1}{2}, 1]) = 0$ while $\mu((\frac{1}{2}, 1]) = \frac{1}{2}$.
The opposite extreme is two measures that never look at the same part of the space. Saying that takes one more word.
$\lambda$ is concentrated on $A \in \mathcal{M}$ if $\lambda(E) = \lambda(E \cap A)$ for every $E \in \mathcal{M}$.
So $\lambda$ lives on $A$: whatever happens outside $A$, $\lambda$ doesn't notice. A measure lives on many sets at once, as long as each one holds all of its mass.
On the Borel sets of $\mathbb{R}$, which of $\{0, 1\}$, $[0, 1]$, $\mathbb{Q}$, and $\{0\}$ is $\lambda = \delta_0 + 2\delta_1$ concentrated on?
$\lambda(E)$ only depends on whether $0$ and $1$ are in $E$. If $0, 1 \in A$, then $E \cap A$ contains $0$ exactly when $E$ does, and the same for $1$, so $\lambda(E \cap A) = \lambda(E)$. So $\lambda$ is concentrated on $\{0, 1\}$, on $[0, 1]$, and on $\mathbb{Q}$. It is not concentrated on $\{0\}$: with $E = \{1\}$, $\lambda(E) = 2$ but $\lambda(E \cap \{0\}) = \lambda(\emptyset) = 0$.
Two measures that live on disjoint sets get a name.
Measures $\lambda_1$ and $\lambda_2$ on $\mathcal{M}$ are mutually singular, written $\lambda_1 \perp \lambda_2$, if there are disjoint sets $A, B \in \mathcal{M}$ such that $\lambda_1$ is concentrated on $A$ and $\lambda_2$ is concentrated on $B$.
Singular has nothing to do with size: $\delta_0 \perp m$ on $\mathbb{R}$, and either one can be scaled up as much as you like. It's only about where they live.
The symbol $\perp$ also means orthogonal on Inner Products & Projections. The two never meet on one page, and the idea is similar: nothing in common. Some notes call this "mutually disjoint"; this site says mutually singular.
Watch both definitions happen to one measure. The density below spreads its mass out, so a tiny set gets a tiny share. Then the same mass gets squeezed onto three points, and the tiny set takes all of it.
Three examples, all against Lebesgue measure $m$ on $[0, 1]$, show that the two definitions are opposite ends of a spectrum, with plenty in between.
On $[0, 1]$, decide whether $\lambda \ll m$, $\lambda \perp m$, or neither, for (a) $d\lambda = x^2 \, dm$, (b) $\lambda = \delta_{1/2}$, (c) $\lambda = m + \delta_{1/2}$.
(a) If $m(E) = 0$, then $\int_E x^2 \, dm = 0$, so $\lambda \ll m$. It is not singular: any set carrying all of $\lambda$ must have full Lebesgue measure, since $x^2 > 0$ off $\{0\}$, so $m$ can't live on a set disjoint from it.
(b) $\lambda$ is concentrated on $\{\frac{1}{2}\}$ and $m$ on $[0, 1] \setminus \{\frac{1}{2}\}$, so $\lambda \perp m$. And $\lambda \not\ll m$: the set $\{\frac{1}{2}\}$ has $m$-measure $0$ and $\lambda$-measure $1$.
(c) Neither. $\{\frac{1}{2}\}$ still witnesses $\lambda \not\ll m$. And $\lambda$ is not singular, because its density part sees exactly what $m$ sees: if $\lambda$ lived on $A$ and $m$ on a disjoint $B$, then $m(A) = m(A \cap B) = 0$, so $\lambda(X \setminus A) \ge m(X \setminus A) = 1$, which contradicts $\lambda$ living on $A$.
Your turn. A witness for $\lambda \not\ll \mu$ is a set with $\mu(E) = 0 < \lambda(E)$. Hunt for them, and notice when there's nothing to find.
A handful of rules for combining these relations will run the rest of the page. The first batch is about singularity.
Let $\lambda, \lambda_1, \lambda_2$ be measures on $\mathcal{M}$ (positive or complex) and $\mu$ positive.
(a) If $\lambda$ is concentrated on $A$, so is $|\lambda|$.
(b) If $\lambda_1 \perp \lambda_2$, then $|\lambda_1| \perp |\lambda_2|$.
(c) If $\lambda_1 \perp \mu$ and $\lambda_2 \perp \mu$, then $\lambda_1 + \lambda_2 \perp \mu$.
Proof. (a) Let $E \in \mathcal{M}$. For a measurable partition $(E_n)$ of $E$, $\lambda(E_n) = \lambda(E_n \cap A)$, and the sets $E_n \cap A$ partition $E \cap A$, so $\sum_n |\lambda(E_n)| = \sum_n |\lambda(E_n \cap A)| \le |\lambda|(E \cap A)$. Taking the sup, $|\lambda|(E) \le |\lambda|(E \cap A)$. The other inequality is monotonicity of the positive measure $|\lambda|$.
(b) If $\lambda_1$ lives on $A$ and $\lambda_2$ on a disjoint $B$, then by (a) $|\lambda_1|$ lives on $A$ and $|\lambda_2|$ on $B$.
(c) Say $\lambda_1$ lives on $A_1$ and $\mu$ on $B_1$, and $\lambda_2$ lives on $A_2$ and $\mu$ on $B_2$, with $A_i \cap B_i = \emptyset$. Then $\mu$ lives on $B = B_1 \cap B_2$, since $\mu(E) = \mu(E \cap B_1) = \mu(E \cap B_1 \cap B_2)$. Each $\lambda_i$ lives on $A = A_1 \cup A_2$, since $\lambda_i(E \setminus A) = \lambda_i((E \setminus A) \cap A_i) = 0$, so $\lambda_1 + \lambda_2$ lives on $A$ too. And $A \cap B = \emptyset$. $\blacksquare$
The second batch mixes the two relations, and ends with the fact that makes every uniqueness claim on this page work: nothing nonzero is both.
Let $\lambda, \lambda_1, \lambda_2$ be measures on $\mathcal{M}$ (positive or complex) and $\mu$ positive.
(d) If $\lambda_1 \ll \mu$ and $\lambda_2 \ll \mu$, then $\lambda_1 + \lambda_2 \ll \mu$.
(e) If $\lambda \ll \mu$, then $|\lambda| \ll \mu$.
(f) If $\lambda_1 \ll \mu$ and $\lambda_2 \perp \mu$, then $\lambda_1 \perp \lambda_2$.
(g) If $\lambda \ll \mu$ and $\lambda \perp \mu$, then $\lambda = 0$.
Proof. (d) If $\mu(E) = 0$, then $\lambda_1(E) + \lambda_2(E) = 0 + 0$.
(e) If $\mu(E) = 0$ and $(E_n)$ is a measurable partition of $E$, then $\mu(E_n) \le \mu(E) = 0$, so $\lambda(E_n) = 0$ for every $n$. Every sum in the definition of $|\lambda|(E)$ is $0$, so $|\lambda|(E) = 0$.
(f) Let $\lambda_2$ live on $A$ and $\mu$ on a disjoint $B$. Then $\mu(A) = \mu(A \cap B) = 0$, so by (e) $|\lambda_1|(A) = 0$. Hence $|\lambda_1(E \cap A)| \le |\lambda_1|(A) = 0$, so $\lambda_1(E) = \lambda_1(E \setminus A)$: $\lambda_1$ lives on $X \setminus A$, which is disjoint from $A$.
(g) By (f) with $\lambda_1 = \lambda_2 = \lambda$, $\lambda$ lives on $X \setminus A$ and on $A$ for some $A$. So $\lambda(E) = \lambda(E \cap A) = \lambda(E \cap A \cap (X \setminus A)) = \lambda(\emptyset) = 0$. $\blacksquare$
Part (g) is the one to remember: a measure that's absolutely continuous and singular with respect to the same $\mu$ is zero. Check that you can tell the two relations apart.
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Here is the core of the chapter. Relative to a σ-finite $\mu$, every complex measure is a density part plus a singular part, with nothing left over.
Let $\mu$ be a positive σ-finite measure and $\lambda$ a complex measure on $(X, \mathcal{M})$.
(a) There is a unique pair of complex measures $\lambda_a$ and $\lambda_s$ on $\mathcal{M}$ such that $$\lambda = \lambda_a + \lambda_s, \qquad \lambda_a \ll \mu, \qquad \lambda_s \perp \mu.$$ (b) There is a unique $h \in L^1(\mu)$ such that $$\lambda_a(E) = \int_E h \, d\mu \quad \text{for every } E \in \mathcal{M}.$$
Proof of uniqueness. Suppose $(\lambda_a', \lambda_s')$ also satisfies (a). Then $\lambda_a' - \lambda_a = \lambda_s - \lambda_s'$. The left side is $\ll \mu$ by part (d) of the proposition above (with $-\lambda_a \ll \mu$), and the right side is $\perp \mu$ by part (c). By part (g), both sides are the zero measure, so $\lambda_a' = \lambda_a$ and $\lambda_s' = \lambda_s$.
If $h, h' \in L^1(\mu)$ both satisfy (b), then $\int_E (h - h') \, d\mu = 0$ for every $E \in \mathcal{M}$. Taking $E = \{\operatorname{Re}(h - h') > 0\}$ gives $\int_E \operatorname{Re}(h - h') \, d\mu = 0$ with a positive integrand on $E$, so $\mu(E) = 0$. Doing the same for $\operatorname{Re}(h - h') < 0$ and for the imaginary part, $h = h'$ $\mu$-a.e. $\blacksquare$
Existence is the hard half, and the next section proves it. First see the pieces once, in the simplest case.
On $[0, 1]$, let $\mu = m$ and $\lambda = 2x \, dm + \delta_{1/2}$. Find $\lambda_a$, $\lambda_s$, and $h$.
The density part is $\ll m$, and the atom lives on $\{\frac{1}{2}\}$, an $m$-null set. So $\lambda_a = 2x \, dm$ and $\lambda_s = \delta_{1/2}$, with $\lambda_s$ concentrated on $\{\frac{1}{2}\}$ and $m$ on $[0, 1] \setminus \{\frac{1}{2}\}$. The function is $h(x) = 2x$, since $\lambda_a(E) = \int_E 2x \, dm$. By uniqueness, no other split works.
Now names for what the theorem produces.
The pair $(\lambda_a, \lambda_s)$ in the theorem is the Lebesgue decomposition of $\lambda$ relative to $\mu$.
If $\lambda \ll \mu$ to begin with, uniqueness forces $\lambda_s = 0$ and $\lambda_a = \lambda$, and then $\lambda(E) = \int_E h \, d\mu$ for every $E$. That's the converse we asked for, and the function earns a name.
The function $h \in L^1(\mu)$ in the theorem is the Radon–Nikodym derivative of $\lambda_a$ with respect to $\mu$, written $h = \frac{d\lambda_a}{d\mu}$ or $d\lambda_a = h \, d\mu$.
$h$ is only pinned down $\mu$-almost everywhere, like any element of $L^1(\mu)$. Here's a decomposition where $\mu$ itself has an atom, which is where people slip.
On $[0, 3]$, let $\mu = m + \delta_2$ and $\lambda = x \, dm + 2\delta_1 + 3\delta_2$. Find $\lambda_a$, $\lambda_s$, and $h = \frac{d\lambda_a}{d\mu}$.
Sort the pieces of $\lambda$. The density part $x \, dm$ is $\ll m$, hence $\ll \mu$. The atom at $1$ sits on $\{1\}$, and $\mu(\{1\}) = 0$, so it is singular. The atom at $2$ sits on a point $\mu$ does see: $\mu(\{2\}) = 1$. So $$\lambda_a = x \, dm + 3\delta_2, \qquad \lambda_s = 2\delta_1.$$ $\lambda_s$ lives on $\{1\}$ and $\mu$ on $[0, 3] \setminus \{1\}$, so $\lambda_s \perp \mu$. For the derivative, take $h(x) = x$ for $x \ne 2$ and $h(2) = 3$. Then $$\int_E h \, d\mu = \int_E h \, dm + h(2)\,\delta_2(E) = \int_E x \, dm + 3\delta_2(E) = \lambda_a(E).$$ The value $h(2) = 3 = \lambda(\{2\})/\mu(\{2\})$ matters, because $\{2\}$ has positive $\mu$-measure.
Here's the decomposition as a picture: the atoms lift off onto a Lebesgue-null set, and the density stays behind.
Which pieces count as singular depends on $\mu$, not on $\lambda$ alone. Build a $\lambda$ and change $\mu$ under it.
The notation $d\lambda = h \, d\mu$ is built to be abused, and the abuse is usually legal. Here's the most useful case.
$\frac{d\lambda}{d\mu}$ looks like a fraction, cancels like a fraction, and is legally not a fraction.
Let $\mu$ be a σ-finite positive measure, $\nu$ a finite positive measure with $\nu \ll \mu$, and $\lambda$ a complex measure with $\lambda \ll \nu$. Prove the chain rule: $\lambda \ll \mu$ and $$\frac{d\lambda}{d\mu} = \frac{d\lambda}{d\nu} \cdot \frac{d\nu}{d\mu} \quad \mu\text{-a.e.}$$
Proof. If $\mu(E) = 0$, then $\nu(E) = 0$, so $\lambda(E) = 0$: $\lambda \ll \mu$. Since $\nu$ is finite (so σ-finite), the theorem gives $f = \frac{d\lambda}{d\nu} \in L^1(\nu)$, and since $\nu \ll \mu$ it gives $g = \frac{d\nu}{d\mu} \in L^1(\mu)$. Here $g \ge 0$ $\mu$-a.e. (if $g < 0$ on a set of positive measure, $\nu$ of that set would be negative), so take $g \ge 0$.
By the density theorem from The Lebesgue Integral, $\int_X \varphi \, d\nu = \int_X \varphi g \, d\mu$ for every measurable $\varphi \ge 0$. Apply it to $\varphi = |f|$: $\int_X |f| g \, d\mu = \int_X |f| \, d\nu < \infty$, so $fg \in L^1(\mu)$. Apply it to $\varphi = (\operatorname{Re} f)^\pm \chi_E$ and $(\operatorname{Im} f)^\pm \chi_E$, which all have finite integrals, and combine the four: $$\lambda(E) = \int_E f \, d\nu = \int_E f g \, d\mu \quad \text{for every } E \in \mathcal{M}.$$ So $fg$ satisfies part (b) for $\lambda$ relative to $\mu$, and by uniqueness $\frac{d\lambda}{d\mu} = fg$ $\mu$-a.e. $\blacksquare$
So the fraction notation tells the truth here, but only because of a theorem. Now compute some derivatives. On a finite space, a point's mass divided by a point's mass is the whole story; with densities, divide the densities.
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And split a measure. Before measuring anything, sort each piece: does $\mu$ see where it lives?
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The proof of existence is due to von Neumann, and it's sneaky: it never constructs $h$ directly. It gets $h$ out of a Hilbert space.
On Inner Products & Projections, every bounded linear functional on a Hilbert space was an inner product: $\Lambda x = \langle x, y \rangle$ for one $y$. On $L^2(\varphi)$, that $y$ is a function.
Start with $\lambda$ and $\mu$ both finite and positive. The plan:
To find $\lambda$'s density, the proof first mashes $\lambda$ and $\mu$ into one measure. Sure. Why not.
On a finite space, every step is one division per point. Watch it run on six points.
Notice the formula: $g = \frac{\lambda}{\mu + \lambda}$, so $\frac{g}{1 - g} = \frac{\lambda}{\mu}$. The detour through $\varphi$ is how the proof divides by $\mu$ without ever dividing by zero. Drag the bars and try it.
On a general space you can't divide point by point, and $g/(1 - g)$ shows up as the sum of a geometric series instead. The proof feeds $f = (1 + g + \cdots + g^n)\chi_E$ into the Riesz identity and lets $n \to \infty$ on both sides.
Here is the whole argument, including the reduction from σ-finite $\mu$ and complex $\lambda$ to the finite positive case.
Let $\mu$ be a positive σ-finite measure and $\lambda$ a complex measure on $(X, \mathcal{M})$. Then there are complex measures $\lambda_a \ll \mu$ and $\lambda_s \perp \mu$ with $\lambda = \lambda_a + \lambda_s$, and an $h \in L^1(\mu)$ with $\lambda_a(E) = \int_E h \, d\mu$ for every $E \in \mathcal{M}$.
Proof. Step 1: a bounded functional. First let $\lambda$ and $\mu$ be finite positive measures, and put $\varphi = \mu + \lambda$. For measurable $f \ge 0$, $\int f \, d\varphi = \int f \, d\mu + \int f \, d\lambda$ (true for characteristic functions, then simple functions, then by monotone convergence). For $f \in L^2(\varphi)$, by the Cauchy–Schwarz inequality, $$\left|\int_X f \, d\lambda\right| \le \int_X |f| \, d\lambda \le \int_X |f| \, d\varphi \le \left(\int_X |f|^2 \, d\varphi\right)^{1/2} \varphi(X)^{1/2} = \|f\|_{L^2(\varphi)} \, \varphi(X)^{1/2}.$$ If $f = 0$ $\varphi$-a.e., then $f = 0$ $\lambda$-a.e. (as $\lambda \le \varphi$), so $\Lambda(f) = \int_X f \, d\lambda$ is a well-defined bounded linear functional on $L^2(\varphi)$.
Step 2: Riesz. By the Riesz representation theorem on a Hilbert space there is $g_0 \in L^2(\varphi)$ with $\Lambda(f) = \langle f, g_0 \rangle = \int_X f \overline{g_0} \, d\varphi$. Put $g = \overline{g_0}$: $$\int_X f \, d\lambda = \int_X f g \, d\varphi \quad \text{for every } f \in L^2(\varphi). \tag{1}$$
Step 3: $0 \le g \le 1$. For $E \in \mathcal{M}$ with $\varphi(E) > 0$, put $f = \chi_E$ in (1): the average of $g$ over $E$ is $$\frac{1}{\varphi(E)} \int_E g \, d\varphi = \frac{\lambda(E)}{\varphi(E)} \in [0, 1],$$ because $0 \le \lambda(E) \le \lambda(E) + \mu(E) = \varphi(E)$. Since $\varphi$ is finite and $[0, 1]$ is closed, the averages theorem from Integrable Functions gives $g(x) \in [0, 1]$ for $\varphi$-a.e. $x$. Changing $g$ on a $\varphi$-null set changes no integral in (1), so take $0 \le g \le 1$ everywhere. Since $\varphi = \mu + \lambda$, (1) rearranges to $$\int_X f (1 - g) \, d\lambda = \int_X f g \, d\mu. \tag{2}$$ It holds for bounded measurable $f$ (those are in $L^2(\varphi)$).
Step 4: the cut. Let $A = \{x : g(x) < 1\}$ and $B = \{x : g(x) = 1\}$, and define $\lambda_a(E) = \lambda(E \cap A)$ and $\lambda_s(E) = \lambda(E \cap B)$. These are positive measures with $\lambda = \lambda_a + \lambda_s$. Put $f = \chi_B$ in (2): since $g = 1$ on $B$, $$\mu(B) = \int_X \chi_B \, g \, d\mu = \int_X \chi_B (1 - g) \, d\lambda = 0.$$ So $\mu$ lives on $A$, $\lambda_s$ lives on $B$, and $\lambda_s \perp \mu$.
Step 5: the geometric series. Fix $E \in \mathcal{M}$ and put $f = (1 + g + \cdots + g^n)\chi_E$ in (2): $$\int_E (1 - g^{n+1}) \, d\lambda = \int_E (g + g^2 + \cdots + g^{n+1}) \, d\mu.$$ On the left, $1 - g^{n+1} = 0$ on $B$ and $1 - g^{n+1} \to 1$ on $A$ (as $0 \le g < 1$ there), with $|1 - g^{n+1}| \le 1$ and $\lambda$ finite, so by dominated convergence the left side tends to $\lambda(E \cap A) = \lambda_a(E)$. On the right the integrands increase to $h = \frac{g}{1 - g}$ on $A$ (and to $\infty$ on $B$, which is $\mu$-null), so by monotone convergence the right side tends to $\int_E h \, d\mu$, with $h = \frac{g}{1-g}$ on $A$ and $h = 0$ on $B$.
Step 6: finishing the finite case. So $\lambda_a(E) = \int_E h \, d\mu$ for every $E$. Taking $E = X$, $\int_X h \, d\mu = \lambda(A) < \infty$, and $h \ge 0$, so $h \in L^1(\mu)$. And $\lambda_a \ll \mu$: if $\mu(E) = 0$ then $\lambda_a(E) = \int_E h \, d\mu = 0$.
Step 7: σ-finite $\mu$, finite positive $\lambda$. Write $X$ as a disjoint union of sets $X_i \in \mathcal{M}$ with $\mu(X_i) < \infty$. Apply Steps 1–6 to $\mu$ and $\lambda$ restricted to subsets of $X_i$, getting $A_i, B_i \subseteq X_i$ and $h_i \ge 0$ on $X_i$. Let $A = \bigcup_i A_i$, $B = \bigcup_i B_i$, $h = h_i$ on $X_i$, $\lambda_a(E) = \lambda(E \cap A)$, $\lambda_s(E) = \lambda(E \cap B)$. Then $\mu(B) = \sum_i \mu(B_i) = 0$, so $\lambda_s \perp \mu$; by monotone convergence $\lambda_a(E) = \sum_i \lambda(E \cap A_i) = \sum_i \int_{E \cap X_i} h_i \, d\mu = \int_E h \, d\mu$; and $\int_X h \, d\mu = \lambda(A) \le \lambda(X) < \infty$.
Step 8: complex $\lambda$. Write $\lambda = \lambda_1 + i\lambda_2$ with real measures $\lambda_1, \lambda_2$, and Jordan-decompose each: $\lambda_k = \lambda_k^+ - \lambda_k^-$, four finite positive measures. Apply Step 7 to each, and combine the four decompositions with the same signs. The combination of the absolutely continuous parts is $\ll \mu$ (part (d)), the combination of the singular parts is $\perp \mu$ (part (c)), and the combination of the four derivatives is in $L^1(\mu)$ and integrates to $\lambda_a$. $\blacksquare$
The theorem has two size hypotheses: $\mu$ is σ-finite, and $\lambda$ is a complex measure, so it's finite. The proof used σ-finiteness once, in Step 7, to chop $X$ into finite pieces. Neither hypothesis is decoration.
The one on $\lambda$ can be relaxed a little: if $\lambda$ is a σ-finite positive measure, the same proof run piece by piece still splits it, though $h$ may no longer be in $L^1(\mu)$. Beyond that, part (a) fails. Take $\mu = m$ and $\lambda$ = counting measure on $(0, 1)$, which is not σ-finite. A singular part would have to live on an $m$-null set $B$, and an absolutely continuous part gives every point measure $0$, since $m(\{x\}) = 0$. So for $x \notin B$, $1 = \lambda(\{x\}) = \lambda_a(\{x\}) + \lambda_s(\{x\}) = 0 + 0$. That forces $B = (0, 1)$, which is not $m$-null. Part (b) fails when $\mu$ is the one that isn't σ-finite.
Let $\mu$ be counting measure on the Borel sets of $[0, 1]$. Show that $m \ll \mu$, but there is no measurable $h\colon [0, 1] \to [0, \infty]$ with $m(E) = \int_E h \, d\mu$ for every Borel $E$.
Proof. $\mu(E) = 0$ only for $E = \emptyset$, and $m(\emptyset) = 0$, so $m \ll \mu$. Suppose $h$ existed. For each $x$, $0 = m(\{x\}) = \int_{\{x\}} h \, d\mu = h(x)\mu(\{x\}) = h(x)$, so $h = 0$. But then $m([0, 1]) = \int_{[0, 1]} 0 \, d\mu = 0$, not $1$. $\blacksquare$
The culprit is $\mu$: counting measure on an uncountable set isn't σ-finite, so the theorem never promised anything.
The name "absolutely continuous" sounds like a continuity condition, and for finite measures it is one: small $\mu$-measure forces small $\lambda$-measure, uniformly.
Let $\mu$ be a positive measure and $\lambda$ a complex measure on $\mathcal{M}$. The following are equivalent:
(1) $\lambda \ll \mu$.
(2) For every $\varepsilon > 0$ there is $\delta > 0$ such that $|\lambda(E)| < \varepsilon$ for every $E \in \mathcal{M}$ with $\mu(E) < \delta$.
Proof. (2) ⇒ (1). If $\mu(E) = 0$, then $\mu(E) < \delta$ for every $\delta > 0$, so $|\lambda(E)| < \varepsilon$ for every $\varepsilon > 0$, and $\lambda(E) = 0$.
(1) ⇒ (2), by contrapositive. Suppose some $\varepsilon > 0$ has no $\delta$. Then for each $n$ there is $E_n \in \mathcal{M}$ with $\mu(E_n) < 2^{-n}$ and $|\lambda(E_n)| \ge \varepsilon$. Let $A_n = \bigcup_{i \ge n} E_i$ and $A = \bigcap_n A_n$. Then $$\mu(A_n) \le \sum_{i=n}^\infty \mu(E_i) < \sum_{i=n}^\infty 2^{-i} = 2^{1-n},$$ and the $A_n$ decrease with $\mu(A_1) < \infty$, so $\mu(A) = \lim_n \mu(A_n) = 0$. On the other hand $|\lambda|(A_n) \ge |\lambda|(E_n) \ge |\lambda(E_n)| \ge \varepsilon$, and $|\lambda|$ is a finite measure (complex measures are bounded), so continuity from above gives $$|\lambda|(A) = \lim_n |\lambda|(A_n) \ge \varepsilon > 0.$$ So $\mu(A) = 0$ but $|\lambda|(A) > 0$: $|\lambda| \not\ll \mu$, and by part (e) of the proposition on absolute continuity, $\lambda \not\ll \mu$. $\blacksquare$
For a nice density you can write $\delta$ down directly.
On $[0, 1]$, let $d\lambda = 2x \, dm$. Given $\varepsilon > 0$, find a $\delta$ that works in (2).
The density is at most $2$, so $|\lambda(E)| = \int_E 2x \, dm \le 2m(E)$. Take $\delta = \frac{\varepsilon}{2}$: if $m(E) < \delta$, then $|\lambda(E)| \le 2m(E) < \varepsilon$. Any bounded density works the same way, with $\delta = \varepsilon / \sup|f|$.
An unbounded density has no such shortcut, and the proposition needs a real proof. That proof leaned on one thing in the last step: $|\lambda|(A_1) < \infty$, so that continuity from above applies. A positive measure can be infinite, and then the proposition breaks.
Play the adversary. You choose $\varepsilon$, the widget picks a $\delta$, and you try to fit more than $\varepsilon$ of $\lambda$ into a set of measure $\delta$.
The middle option is worth a second look: its density is unbounded, but its total mass is finite, and that's all the proposition needs. The last option has infinite mass. Prove that it really breaks.
On $(0, 1)$, let $\lambda(E) = \int_E \frac{1}{x} \, dm$. Show that $\lambda$ is a positive measure with $\lambda \ll m$, but that (2) fails: for $\varepsilon = 1$, no $\delta > 0$ works.
Proof. $\lambda$ is a positive measure by the density theorem from The Lebesgue Integral, and if $m(E) = 0$, then $\int_E \frac{1}{x} \, dm = 0$, so $\lambda \ll m$. Given any $\delta > 0$, let $E = (0, \frac{\delta}{2})$. Then $m(E) = \frac{\delta}{2} < \delta$, but $$\lambda(E) = \int_0^{\delta/2} \frac{dx}{x} = \lim_{t \to 0^+} \big(\ln \tfrac{\delta}{2} - \ln t\big) = \infty \ge 1.$$ So no $\delta$ works for $\varepsilon = 1$. $\blacksquare$
One family of complex measures comes up constantly: $E \mapsto \int_E f \, d\mu$ for $f \in L^1(\mu)$. Each one is $\ll \mu$, so each has its own $\delta$. Asking for one $\delta$ that works for a whole family of functions at once has a name.
Let $(X, \mathcal{M}, \mu)$ be a positive measure space. A set $\Phi \subseteq L^1(\mu)$ is uniformly integrable if for every $\varepsilon > 0$ there is $\delta > 0$ such that $\left|\int_E f \, d\mu\right| < \varepsilon$ for every $f \in \Phi$ and every $E \in \mathcal{M}$ with $\mu(E) < \delta$.
A single function is uniformly integrable by the proposition. Finitely many are too.
Prove that every finite subset of $L^1(\mu)$ is uniformly integrable.
Proof. Let $\Phi = \{f_1, \ldots, f_N\} \subseteq L^1(\mu)$ and $\varepsilon > 0$. For each $k$, $\lambda_k(E) = \int_E f_k \, d\mu$ is a complex measure (countable additivity is dominated convergence, since $f_k \in L^1(\mu)$), and $\lambda_k \ll \mu$ because an integral over a null set is $0$. By the proposition there is $\delta_k > 0$ with $|\lambda_k(E)| < \varepsilon$ whenever $\mu(E) < \delta_k$. Let $\delta = \min\{\delta_1, \ldots, \delta_N\} > 0$. If $\mu(E) < \delta$, then $\left|\int_E f_k \, d\mu\right| < \varepsilon$ for every $k$. $\blacksquare$
Infinite families are where it gets interesting, and the Spicy Bonus at the bottom of the page shows what uniform integrability buys you.
Now point the theorem at a single complex measure $\mu$ and its own total variation. Since $|\mu(E)| \le |\mu|(E)$, every $|\mu|$-null set is $\mu$-null: $\mu \ll |\mu|$. So $\mu$ has a density with respect to $|\mu|$, and that density turns out to have modulus $1$.
Let $\mu$ be a complex measure on $\mathcal{M}$. Then there is a measurable $h$ with $|h(x)| = 1$ for every $x \in X$ such that $$d\mu = h \, d|\mu|.$$
Proof. $|\mu|$ is a finite positive measure and $\mu \ll |\mu|$, so the Radon–Nikodym theorem gives $h \in L^1(|\mu|)$ with $d\mu = h \, d|\mu|$.
$|h| \ge 1$ a.e. Let $A_n = \{x : |h(x)| < 1 - \frac{1}{n}\}$. For a measurable partition $(E_k)$ of $A_n$, $$\sum_k |\mu(E_k)| = \sum_k \left|\int_{E_k} h \, d|\mu|\right| \le \sum_k \left(1 - \tfrac{1}{n}\right)|\mu|(E_k) = \left(1 - \tfrac{1}{n}\right)|\mu|(A_n).$$ Taking the sup over partitions, $|\mu|(A_n) \le (1 - \frac{1}{n})|\mu|(A_n)$, which forces $|\mu|(A_n) = 0$. So $|\mu|(\{|h| < 1\}) = |\mu|\big(\bigcup_n A_n\big) = 0$.
$|h| \le 1$ a.e. If $|\mu|(E) > 0$, the average of $h$ over $E$ is $$\frac{1}{|\mu|(E)} \int_E h \, d|\mu| = \frac{\mu(E)}{|\mu|(E)},$$ a complex number in the closed unit disk, since $|\mu(E)| \le |\mu|(E)$. The disk is closed, so by the averages theorem $|h| \le 1$ $|\mu|$-a.e.
So $|h| = 1$ outside a $|\mu|$-null set. Redefine $h = 1$ on that set; this changes no integral against $|\mu|$. $\blacksquare$
Read $d\mu = h \, d|\mu|$ like the polar form $z = e^{i\theta}|z|$ of a complex number: $|\mu|$ says how much mass sits where, and $h$ says which way it points. That settles a promise from the last page.
On Total Variation, a real density had $|f \, d\lambda| = |f| \, d\lambda$. The same is true for complex densities, and the polar decomposition is the tool.
Let $\mu$ be a positive measure, $g \in L^1(\mu)$, and $\lambda(E) = \int_E g \, d\mu$. Then $$|\lambda|(E) = \int_E |g| \, d\mu \quad \text{for every } E \in \mathcal{M}.$$
Proof. $\lambda$ is a complex measure since $g \in L^1(\mu)$ (countable additivity is dominated convergence). By the polar decomposition, $d\lambda = h \, d|\lambda|$ with $|h| = 1$. So for every $E \in \mathcal{M}$, $$\int_X \chi_E \, h \, d|\lambda| = \lambda(E) = \int_X \chi_E \, g \, d\mu.$$ By linearity the same holds with $\chi_E$ replaced by any simple function. A bounded measurable $f$ is a pointwise limit of simple functions $s_n$ with $|s_n| \le 2\sup|f|$, and since $h \in L^1(|\lambda|)$ and $g \in L^1(\mu)$, dominated convergence gives $\int_X f h \, d|\lambda| = \int_X f g \, d\mu$. Take $f = \overline{h}\chi_E$. Since $\overline{h} h = |h|^2 = 1$, $$|\lambda|(E) = \int_E \overline{h} g \, d\mu \quad \text{for every } E \in \mathcal{M}.$$ The left side is real and $\ge 0$ for every $E$. With $E = \{\operatorname{Im}(\overline{h} g) > 0\}$, the imaginary part of the right side is the integral of a positive function over $E$ and must be $0$, so $\mu(E) = 0$; likewise for $\operatorname{Im}(\overline{h} g) < 0$ and for $\operatorname{Re}(\overline{h} g) < 0$. So $\overline{h} g$ is real and $\ge 0$ $\mu$-a.e., and a complex number of that kind equals its modulus: $\overline{h} g = |\overline{h}| \, |g| = |g|$ $\mu$-a.e. Hence $|\lambda|(E) = \int_E |g| \, d\mu$. $\blacksquare$
Watch the cancellation that $|\lambda|$ refuses to do.
On $[0, 2]$, let $d\lambda = g \, dm$ with $g = 3 + 4i$ on $[0, 1)$ and $g = -2$ on $[1, 2]$. Find $h$ in $d\lambda = h \, d|\lambda|$, $|\lambda|([0, 2])$, and $|\lambda([0, 2])|$.
By the corollary, $d|\lambda| = |g| \, dm$, with $|g| = 5$ on $[0, 1)$ and $|g| = 2$ on $[1, 2]$. Then $d\lambda = g \, dm = \frac{g}{|g|} \, |g| \, dm$, so $$h = \frac{g}{|g|} = \begin{cases} \frac{3}{5} + \frac{4}{5}i & \text{on } [0, 1) \\ -1 & \text{on } [1, 2]. \end{cases}$$ The total variation adds lengths: $|\lambda|([0, 2]) = 5 \cdot 1 + 2 \cdot 1 = 7$. The measure itself adds arrows: $\lambda([0, 2]) = (3 + 4i) - 2 = 1 + 4i$, so $|\lambda([0, 2])| = \sqrt{17} \approx 4.12$, which is less than $7$.
The gap between $7$ and $\sqrt{17}$ is exactly the cancellation between the two pieces, and it's the most common slip on this page.
$|\lambda|(E) = \int_E |g| \, d\mu$, not $\left|\int_E g \, d\mu\right|$. The second one is $|\lambda(E)|$, and it lets the pieces cancel. The modulus goes inside the integral.
Compute a polar part, or a total variation.
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Now let $\mu$ be a real measure. Its polar part $h$ is real with $|h| = 1$, so $h = \pm 1$. The set where $h = 1$ holds the positive mass, and the set where $h = -1$ holds the negative mass. The Jordan decomposition was a split of the measure; this is a split of the space.
Let $\mu$ be a real measure on $\mathcal{M}$. Then there are disjoint $A, B \in \mathcal{M}$ with $A \cup B = X$ such that $$\mu^+(E) = \mu(E \cap A) \quad \text{and} \quad \mu^-(E) = -\mu(E \cap B) \quad \text{for every } E \in \mathcal{M}.$$ Such a pair $(A, B)$ is a Hahn decomposition of $X$ for $\mu$.
Proof. By the polar decomposition, $d\mu = h \, d|\mu|$ with $|h| = 1$. Since $\mu$ is real, $h$ is real $|\mu|$-a.e. (its imaginary part integrates to $0$ over every set), so after changing $h$ on a $|\mu|$-null set, $h = \pm 1$ everywhere. Let $A = \{h = 1\}$ and $B = \{h = -1\}$. Then $\frac{1}{2}(1 + h)$ is $1$ on $A$ and $0$ on $B$, so for every $E$, $$\mu^+(E) = \tfrac{1}{2}\big(|\mu|(E) + \mu(E)\big) = \int_E \tfrac{1}{2}(1 + h) \, d|\mu| = |\mu|(E \cap A) = \int_{E \cap A} h \, d|\mu| = \mu(E \cap A).$$ Finally $\mu^-(E) = \mu^+(E) - \mu(E) = \mu(E \cap A) - \big(\mu(E \cap A) + \mu(E \cap B)\big) = -\mu(E \cap B)$. $\blacksquare$
For a measure with a density, the Hahn sets are where the density is positive and where it's negative. Watch the cut, and then a set $E$ sliding across it.
Every real measure has a happy place $A$ and a sad place $B$. Relatable.
Draw your own density and watch $A$ and $B$ follow it. When $f$ is zero on a whole stretch, that stretch can go in either set.
There's a second way to find $A$: it's the set that makes $\mu$ as large as possible. Since $\mu(E) = \mu^+(E) - \mu^-(E) \le \mu^+(X) = \mu(A)$, no set beats $A$. Try to tie it.
Here is a Hahn decomposition worked out for a polynomial density.
On $X = [0, 3]$, let $d\mu = (x - 1) \, dm$. Find a Hahn decomposition, $\mu^+(X)$, and $\mu^-(X)$.
The density $x - 1$ is $\ge 0$ on $[1, 3]$ and $< 0$ on $[0, 1)$, so $A = [1, 3]$ and $B = [0, 1)$ is a Hahn decomposition: they are disjoint, cover $X$, and $\mu$ is positive on subsets of $A$ and negative on subsets of $B$. Then $$\mu^+(X) = \mu(A) = \int_1^3 (x - 1) \, dx = 2, \qquad \mu^-(X) = -\mu(B) = -\int_0^1 (x - 1) \, dx = \tfrac{1}{2}.$$ Check: $\mu(X) = \int_0^3 (x - 1) \, dx = \frac{3}{2} = 2 - \frac{1}{2}$. The point $1$, where the density is $0$, could have gone into $B$ instead: it's a null set.
That last remark is the general truth. A Hahn decomposition is not unique, but only null sets can move.
Let $\mu$ be a real measure with Hahn decompositions $(A, B)$ and $(A', B')$. Prove that $|\mu|(A \cap B') = 0$ and $|\mu|(A' \cap B) = 0$, so $A$ and $A'$ differ by a $|\mu|$-null set.
Proof. Let $S = A \cap B'$. Using $(A', B')$: $\mu^+(S) = \mu(S \cap A') = \mu(\emptyset) = 0$, since $S \subseteq B'$ is disjoint from $A'$. Using $(A, B)$: $\mu^-(S) = -\mu(S \cap B) = -\mu(\emptyset) = 0$, since $S \subseteq A$ is disjoint from $B$. So $|\mu|(S) = \mu^+(S) + \mu^-(S) = 0$. Swapping the two decompositions gives $|\mu|(A' \cap B) = 0$. Since $A \setminus A' = A \cap B'$ and $A' \setminus A = A' \cap B$, the symmetric difference of $A$ and $A'$ is $|\mu|$-null. $\blacksquare$
The Hahn decomposition also says the Jordan decomposition is the most economical way to write $\mu$ as a difference of positive measures.
If $\mu = \lambda_1 - \lambda_2$, where $\lambda_1$ and $\lambda_2$ are finite positive measures, then $\mu^+ \le \lambda_1$ and $\mu^- \le \lambda_2$.
Proof. Let $(A, B)$ be a Hahn decomposition for $\mu$. For every $E \in \mathcal{M}$, since $\lambda_2 \ge 0$, $$\mu^+(E) = \mu(E \cap A) = \lambda_1(E \cap A) - \lambda_2(E \cap A) \le \lambda_1(E \cap A) \le \lambda_1(E).$$ In the same way, $\mu^-(E) = -\mu(E \cap B) = \lambda_2(E \cap B) - \lambda_1(E \cap B) \le \lambda_2(E)$. $\blacksquare$
The Total Variation page proved this same corollary as an Exercise, by bounding $|\mu|$ with partitions. With $A$ and $B$ in hand, it's a two-line inequality.
Any other split carries extra mass on both sides that cancels.
Write $\mu = \delta_0 - \delta_1$ as $\lambda_1 - \lambda_2$ with $\lambda_1 = \delta_0 + \delta_2$ and $\lambda_2 = \delta_1 + \delta_2$. Compare with $\mu^+$ and $\mu^-$.
A Hahn decomposition is $A = \{0\}$, $B = \mathbb{R} \setminus \{0\}$, so $\mu^+ = \delta_0$ and $\mu^- = \delta_1$. Indeed $\mu^+ = \delta_0 \le \delta_0 + \delta_2 = \lambda_1$ and $\mu^- = \delta_1 \le \lambda_2$. The extra $\delta_2$ on both sides cancels in $\lambda_1 - \lambda_2$, and the Jordan decomposition is the split without it.
Last lesson question: find the Hahn sets, or the mass on one side of them.
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On Duality: (Lᵖ)* and C₀(X)*, the next page, a bounded functional $\Lambda$ on $L^p(\mu)$ becomes a measure $E \mapsto \Lambda(\chi_E)$. That measure is $\ll \mu$, and its Radon–Nikodym derivative is the function $g \in L^q$ that represents $\Lambda$.
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. Uniform integrability looked like a technicality. Combined with Egoroff's theorem, it gives a convergence theorem with no dominating function at all.
Prove Vitali's convergence theorem. Let $\mu(X) < \infty$, let $\{f_n\} \subseteq L^1(\mu)$ be uniformly integrable, and suppose $f_n(x) \to f(x)$ for $\mu$-a.e. $x$, with $|f(x)| < \infty$ a.e. Then $f \in L^1(\mu)$ and $$\lim_{n \to \infty} \int_X |f_n - f| \, d\mu = 0.$$
Proof. Changing every $f_n$ and $f$ on one null set changes no integral, so assume $f_n(x) \to f(x) \in \mathbb{C}$ for every $x$. Then $f$ is measurable, as a pointwise limit of measurable functions.
Step 1: control of $\int_E |f_n|$. Let $\varepsilon > 0$ and take $\delta$ from uniform integrability for $\varepsilon$. Let $\mu(E) < \delta$. With $P = E \cap \{\operatorname{Re} f_n > 0\}$ and $N = E \cap \{\operatorname{Re} f_n < 0\}$, both of measure $< \delta$, $$\int_E |\operatorname{Re} f_n| \, d\mu = \operatorname{Re} \int_P f_n \, d\mu - \operatorname{Re} \int_N f_n \, d\mu \le \left|\int_P f_n \, d\mu\right| + \left|\int_N f_n \, d\mu\right| < 2\varepsilon.$$ The same holds for $\operatorname{Im} f_n$, and $|f_n| \le |\operatorname{Re} f_n| + |\operatorname{Im} f_n|$, so $$\int_E |f_n| \, d\mu < 4\varepsilon \quad \text{for every } n \text{ whenever } \mu(E) < \delta. \tag{$*$}$$
Step 2: Egoroff. Since $\mu(X) < \infty$, Egoroff's theorem gives a measurable $G$ with $\mu(X \setminus G) < \delta$ such that $f_n \to f$ uniformly on $G$. Let $E = X \setminus G$.
Step 3: $f \in L^1(\mu)$. By Fatou's lemma and $(*)$, $\int_E |f| \, d\mu \le \liminf_n \int_E |f_n| \, d\mu \le 4\varepsilon$. On $G$, uniform convergence gives an $N$ with $|f - f_N| \le 1$ on $G$, so $\int_G |f| \, d\mu \le \int_G |f_N| \, d\mu + \mu(G) < \infty$, as $f_N \in L^1(\mu)$ and $\mu(X) < \infty$. So $\int_X |f| \, d\mu < \infty$.
Step 4: the limit. For every $n$, $$\int_X |f_n - f| \, d\mu \le \int_G |f_n - f| \, d\mu + \int_E |f_n| \, d\mu + \int_E |f| \, d\mu \le \mu(X) \sup_G |f_n - f| + 4\varepsilon + 4\varepsilon.$$ The supremum tends to $0$, so $\limsup_n \int_X |f_n - f| \, d\mu \le 8\varepsilon$. Since $\varepsilon$ was arbitrary, the limit is $0$. $\blacksquare$