Source of most content: Gunhee Cho and RCA by Rudin.

Converse of Radon-Nikodym theorem

\[\text{Given a M.S. } (X, F, \mu), \text{ let } f: X \to [0, \infty] \text{ be measurable function. Define } \varphi: F \to [0, \infty] (E\in F \mapsto \int_E f \mathrm{d}\mu). \nonumber\] \[\text{Then, } \varphi \text{ is a measure. Moreover, given a measurable function } g : X \to [0, \infty], \int_X g \mathrm{d}\varphi = \int_X g f \mathrm{d}\mu. \nonumber\]

Proof

• Let $E_1, E_2, \ldots$ be disjoint members of $F$ whose union is $E$. Then, $f(x) \chi_{E}(x) = \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x)$ holds. Therefore, $\varphi (E) = \int_{E} f \mathrm{d}\mu = \int_{X} f \chi_E \mathrm{d}\mu= \int_{X} \sum_{k=1}^{\infty} f(x) \chi_{E_k}(x) \mathrm{d}\mu \stackrel{f \geq 0}{=} \sum_{k=1}^{\infty} \int_{X} f(x) \chi_{E_k}(x) \mathrm{d}\mu = \sum_{k=1}^{\infty} \varphi (E_{k})$ holds. This implies that $\varphi$ satisfies contunable additivity. Also, as $\varphi(\emptyset) = \int_{\emptyset} f\mathrm{d}\mu =0 $, $\varphi$ is a measure.
• NTS: $\int_{X}g \mathrm{d}\varphi = \int_{X} gf \mathrm{d}\mu$.
  1. When $g=\chi_E$ for $E\in F$: $\int_{X} \chi_E \mathrm{d}\varphi = \int_{E} 1 \mathrm{d}\varphi = \varphi(E) = \int_{E} f \mathrm{d}\mu = \int_{X} \chi_{E} f \mathrm{d}\mu$.
  2. When $g$ is simple: $\int_{X}g \mathrm{d}\varphi = \int_{X} \sum_{i} c_{i} \chi_{A_{i}} \mathrm{d}\varphi = \sum_{i} c_{i} \int_{X} \chi_{A_{i}} \mathrm{d}\varphi \stackrel{1}{=} \sum_{i} c_{i} \int_{X} \chi_{A_{i}} f \mathrm{d}\varphi =\int_{X} gf \mathrm{d}\mu$.
  3. When $g$ is non-negative measurable: By simple approximation thm, there exists a monotonic increaseing seq. of non-negative measurable simple functions $ \{ s _{i} \} _{i=1}^{\infty} $ . Then, by 2, $\int _{X} s _i \mathrm{d}\varphi = \int _{X} s _i f \mathrm{d}\varphi$ holds for all $i$. By MCT, $\int _{X}g \mathrm{d}\varphi = \int _{X} gf \mathrm{d}\mu$ holds.

NB1; From this theorem, we can see the integral as a measure.

$L^1$ space

Def. Given a M.S. $(X, F, \mu)$, $L^1$ space $L^1(X, F, \mu) = L^1 (\mu) := \{ f: X \to \mathbb{C} : \int_X | f | \mathrm{d}\mu < \infty \}$ is a collection of Lebesgue integrable functions.

NB2; If $f\in L^1(\mu): X \to \mathbb{C}$, i.e., $f = u + iv$ for real measurable functions $u,v$ on $X$, then $\int_{E} f \mathrm{d}\mu = \int_{E} u^+ \mathrm{d}\mu + \int_{E} u^- \mathrm{d}\mu + i \int_{E} v^+ \mathrm{d}\mu + i \int_{E} v^- \mathrm{d}\mu$. Note that $\int_{E} u^+ \mathrm{d}\mu \leq \int_{E} \mid u \mid \mathrm{d}\mu \leq \int_{E} \mid f \mid \mathrm{d}\mu < \infty$.

Proposition

\[\text{Suppose } f,g \in L^1(\mu), \alpha, \beta \in \mathbb{C}. \text{ Then, } \alpha f + \beta g \in L^1(\mu) \text{ and } \int_{X} \alpha f + \beta g \mathrm{d}\mu = \alpha \int_{X} f \mathrm{d}\mu + \beta \int_{X} g \mathrm{d}\mu. \nonumber\]

Proof

• $\int_{X} \mid \alpha f + \beta g \mid \mathrm{d}\mu \leq \int_{X} \mid \alpha \mid \, \mid f \mid + \mid \beta \mid \, \mid g \mid \mathrm{d}\mu = \mid \alpha \mid \int_{X} \mid f \mid \mathrm{d}\mu + \mid \beta \mid \int_{X} \mid g \mid \mathrm{d}\mu < \infty$.
• 1. Let $h = f+g$. Then, we have $h^+ +f^- + g^- = f^+ + g^+ + g^+ + h^-$. From previous proposition, it holds that $\int h^+ + \int f^- + \int g^- = \int f^+ + \int g^+ + \int h^-$. Since $f$ and $g$ are Lebesgue intergrable, $\int f+g = \int f +\int g$ holds.
  2. NTS: $\int \alpha f \mathrm{d}\mu = \alpha \int f \mathrm{d}\mu$ for $\alpha \in \mathbb{C}$. Since it holds for $\alpha \in \mathbb{R}$, it suffices to show when $\alpha = i$. If $f = u + iv$, then $\int if = \int iu -v = \int -v + i \int u = i(\int u + i \int v) = i \int f $.

---

\[\text{If } f \in L^1(\mu), \text{ then it holds that } \mid \int_X f \mathrm{d}\mu \mid \leq \int_X \mid f \mid \mathrm{d}\mu. \nonumber\]

Proof

Let $z= \int_{X} f \mathrm{d}\mu$ and $\angle z = \theta$. Take $\alpha = e^{i\theta}$, then $\alpha z = \mid z \mid$. Then, $\mid \int_X f \mathrm{d}\mu \mid = \mid z \mid = \alpha z = \int_X Re(f) \mathrm{d}\mu \leq \int_X \mid f \mid \mathrm{d}\mu$ holds.

Lebesgue’s DCT (LDCT)

\[\text{Supppose } \{ f_n \} \text{ is a seq. of complex measurable fns on } X \text{ s.t. } f(x) = \lim_{n\to \infty} f_n(x) \text{ exists } \forall x \in X. \nonumber\] \[\text{If } \exists g \in L^1(\mu) \text{ s.t. } \mid f_n(x) \mid \leq g(x), \text{ then } f\in L^1(\mu), \lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0 \text{ and } \lim_{n\to \infty} \int_X f_n \mathrm{d}\mu = \int_X f \mathrm{d}\mu. \nonumber\]

Proof

First, $\mid f_n \mid \leq g$ gives $\mid f \mid \leq g$, i.e., $f \in L^1 (\mu)$. Since $\mid f_n - f \mid \leq 2g$ and $2g - \mid f_n -f \mid$ is measurable, applying Fatou’s Lemma, we have $\int_X \liminf 2g- \mid f_n - f\mid \mathrm{d}\mu = \int_X 2g \mathrm{d}\mu \leq \liminf \int_X 2g- \mid f_n - f\mid \mathrm{d}\mu $ $= \int_X 2g \mathrm{d}\mu + \liminf_{n\to \infty} \left( -\int_X \mid f_n - f \mid \mathrm{d}\mu \right) = \int_X 2g \mathrm{d}\mu - \limsup \int_X \mid f_n - f \mid \mathrm{d}\mu$. It implies that $\limsup \int_X \mid f_n - f \mid \leq 0$, i.e., $\limsup \int_X \mid f_n - f \mid \leq =0$. Since $\liminf \int_X \mid f_n - f \mid \geq 0$, $\lim_{n\to \infty} \int_X \mid f_n - f \mid \mathrm{d}\mu = 0$ holds. Applying above proposition concludes the proof.

---

NB1; Define $f_n(x) = \frac{f(x+1/n)-f(x)}{1/n}$. Then, $f_n(x) \to f’(x)$ if $f$ is differentiable. If there exists $g\in L^1(\mu)$ s.t. $\forall n, \, \mid f_n \mid \leq g$, then $\frac{d}{dx}\int_X f \mathrm{d}\mu =\lim_n \int_X f_n \mathrm{d}\mu \stackrel{LDCT}{=} \int_X \lim_n f_n \mathrm{d}\mu = \int_X f’ \mathrm{d}\mu = \int_X \frac{d}{dx}f \mathrm{d}\mu$ holds. This shows that we can switch the derivative and integration.