If \(z_1,\ldots,z_N\in \mathbb C\), then there is \(S \subseteq \{1,\ldots,N\}\) such that \[ \Big|\sum_{k \in S}z_k\Big| \geq \frac{1}{\pi}\sum_{k=1}^N|z_k|.\]
Proof. Let \({\color{blue}z_k=e^{i\theta_k}|z_k|}\), and \(S(\theta)=\{k \in \{1,\ldots,N\}: \cos(\theta_k-\theta)>0\}\) for \(-\pi \leq \theta \leq \pi\). Then \[\Big|\sum_{k \in S(\theta)}z_k\Big|=\Big|\sum_{k \in S(\theta)}e^{-i\theta}z_k\Big| \geq {\rm Re} \sum_{k \in S(\theta)}e^{-i\theta}z_k=\sum_{k=1}^{N}|z_k|\cos^{+}(\theta_k-\theta).\]
Choose \(\theta=\theta_0\in [-\pi,\pi]\) to maximize the left sum and set \(S=S(\theta_0)\), then \[\sum_{k=1}^N |z_k|\cos^+(\theta_k-\theta_0) \geq \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{k=1}^{N}|z_k|\cos^{+}(\theta_k-\theta)d\theta=\frac{1}{\pi}\sum_{k=1}^n|z_k|,\] because \(\frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{+}(\theta_k-\theta)d\theta=\frac{1}{\pi}\). This proves the lemma.$$\tag*{$\blacksquare$}$$
If \(\mu\) is a complex measure on \((X,\mathcal M)\), then \(|\mu|(X)<\infty.\)
Proof. Suppose \(|\mu|(E)=\infty\) for some \(E \in \mathcal M\). Put \(t=\pi(1+|\mu(E)|)\) since \(|\mu|(E)>t\). There is a partition \((E_j)_{j \in \mathbb{N}}\subseteq \mathcal M\) of \(E\) such that \[\sum_{i=1}^N |\mu(E_i)|>t \quad \text{ for some } \quad N \in \mathbb{N}.\]
By the previous lemma with \(z_i=\mu(E_i)\) we obtain that there is \(A \subseteq E\) and \[|\mu(A)|>\frac{t}{\pi}>1.\] Setting \(B = E \setminus A\) it follows that \[|\mu(B)|=|\mu(E)-\mu(A)| \geq |\mu(A)|-|\mu(E)|>\frac{t}{\pi}-|\mu(E)|=1.\]
We have split \(E=A \cup B\), into disjoint sets, where \(|\mu(A)|>1\) and \(|\mu(B)|>1\). Then \(|\mu|(A)=\infty\) or \(|\mu|(B)=\infty\), since \(|\mu|\) is a measure and \(|\mu|(E)=\infty\).
Since \(|\mu|(X)=\infty\), split \(X\) into \(A_1\) and \(B_1\) as above with \(|\mu(A_1)|>1\) and \(|\mu|(B_1)=\infty\). Next, we split \(B_1\) into \(A_2\), \(B_2\) with \(|\mu(A_2)|>1\) and \(|\mu|(B_2)=\infty\) and so on.
We obtain a countable disjoint collection \((A_j)_{j \in \mathbb{N}}\subseteq \mathcal M\) such that \[\mu\Big(\bigcup_{j \in \mathbb{N}}A_j\Big)=\sum_{j \in \mathbb{N}}\mu(A_j)\] by the countable additivity of \(\mu\).
This implies that \(\lim_{n \to \infty}\mu(A_n)=0\), which is impossible, since \(|\mu(A_j)|>1\) for all \(j \in \mathbb{N}\). Thus we must have \(|\mu|(X)<\infty\). $$\tag*{$\blacksquare$}$$
Let \(\nu\) be a complex measure (\(\sigma\)-finite signed measure) and \(\mu\) be a \(\sigma\)-finite positive measure on \((X,\mathcal M)\). Then the following are true.
(Lebesgue decomposition) There exist unique complex measures (\(\sigma\)-finite signed measure) \(\nu_a\) and \(\nu_s\) on \((X,\mathcal M)\) such that \[\nu_a\ll \mu \quad \text{ and } \quad \nu_s\perp \mu \quad \text{ and } \quad \nu=\nu_a+\nu_s.\]
(Radon–Nikodym theorem) Moreover, there is \(f\in L^1(X, \mu)\) (there is an extended \(\mu\)-measurable function \(f:X \to \mathbb{R}\)) such that \[\nu_a(E)=\int_Ef(x) d\mu(x) \quad \text{ for } \quad E\in\mathcal M,\] and any two such functions are equal \(\mu\)-a.e. on \(X\).
The decomposition \(\nu=\lambda+\rho\), where \[\lambda \perp \mu \quad \text{ and } \quad \rho \ll \mu\] is called the Lebesgue decomposition of \(\nu\) with respect to \(\mu\).
If \(\nu \ll \mu\) by the Radon–Nikodym theorem we obtain \(d\nu=f\,d\mu\) and \(f\) is called the Radon–Nikodym derivative of \(\nu\) with respect to \(\mu\).
The Radon–Nikodym derivative is sometimes denoted by \[f=\frac{d\nu}{d\mu},\quad \text{ i.e. } \quad d\nu=\frac{d\nu}{d\mu}d\mu.\]
More precisely, \(\frac{d\nu}{d\mu}\) should be constructed as the class of of functions equal to \(f\) \(\mu\)-a.e.
Proposition. Suppose that \(\nu\) is a \(\sigma\)-finite signed measure and \(\mu,\lambda\) are \(\sigma\)-finite measures of \((X,\mathcal M)\) such that \(\nu \ll \mu\) and \(\mu \ll \lambda\).
If \(g \in L^1(X, \nu)\), then \(g\frac{d\nu}{d\mu} \in L^1(X, \mu)\) and \[\int_X g\,d\nu=\int_X g \frac{d\nu}{d\mu}\,d\mu.\]
We have \(\nu \ll \lambda\) and \[\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda},\] holds \(\lambda\)-a.e. on \(X\)
In particular, by (b) we may conclude the following: if \(\mu \ll \lambda\) and \(\lambda \ll \mu\), then \(\frac{d\lambda}{d\mu}\frac{d\mu}{d\lambda}=1\) holds a.e. on \(X\) with respect to either \(\mu\) or \(\lambda\).
Proof. By considering \(\nu^+\) and \(\nu^-\) we may assume that \(\nu \geq 0\).
The equation is true when \(g=\mathbf{1}_{{E}}\) and \(E \in\mathcal M\) since \[\int_E \,d\nu=\int_E\frac{d\nu}{d\mu}\,d\mu\] by the Radon–Nikodym theorem.
It remains true for simple functions by linearity, then for nonnegative measurable functions by the (MCT), and finally for function in \(L^1(X, \nu)\) by linearity again.
Note that \(\nu\ll \lambda\) and \[\int_E \frac{d\nu}{d\lambda}\,d\lambda=\nu(E)=\int_E \frac{d\nu}{d\mu}\,d\mu=\int_E \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\,d\lambda,\] by applying the previous part with \(g=\mathbf{1}_{{E}} \frac{d\nu}{d\mu}\) for all \(E \in \mathcal M\).
Thus \(\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\) \(\lambda\)-a.e. on \(X\) as desired. $$\tag*{$\blacksquare$}$$
Let \(\mu\) be a complex measure on \((X,\mathcal M)\). Then there is a measurable function \(h:X \to \mathbb{C}\) such that \(|h(x)|=1\) for all \(x \in X\) and \[d\mu=f\,d|\mu|.\]
Proof. Since \(\mu \ll |\mu|\) so by the Radon–Nikodym theorem there is \(h \in L^1(X, |\mu|)\) such that \({\color{purple}d\mu=h\,d|\mu|}.\) Let \[A_r=\{x \in X:|h(x)|<r\} \quad \text{ for } \quad r>0.\]
Let \((E_j)_{j \in \mathbb{N}} \subseteq \mathcal M\) be such that \(A_r=\bigcup_{j \in \mathbb{N}}E_j\), then \[\sum_{j \in \mathbb{N}}|\mu(E_j)|=\sum_{j \in \mathbb{N}}\Big|\int_{E_j}h\,d|\mu|\Big| \leq \sum_{j \in \mathbb{N}}r|\mu|(E_j)=r|\mu|(A_r)\] and \(|\mu|(A_r) \leq r|\mu|(A_r)\), giving \(|\mu|(A_r)=0\) if \(r<1\). Thus \(|h| \geq 1\) \(\mu\)-a.e.
If \(|\mu|(E)>0\), by using \({\color{purple}d\mu=h\,d|\mu|}\), we obtain \[\bigg|\frac{1}{|\mu|(E)}\int_E h\,d|\mu|\bigg|=\frac{|\mu(E)|}{|\mu|(E)}\le 1.\]
Proposition (See previous lecture). Suppose that \((X,\mathcal M, \mu)\) is a finite measure space \(\mu(X)<\infty\). Let \(f \in L^1(X)\) and \(S\subseteq \mathbb C\) be a closed set and \[A_Ef=\frac{1}{\mu(E)}\int_Efd\mu \in S\] for every \(E \in \mathcal M\) with \(\mu(E)>0\). Then \(f(x) \in S\) for almost all \(x \in X\).
By this proposition with the closed unit disc in place of \(S\), we conclude that \(|h| \leq 1\) \(\mu\)-a.e. Hence \(|\mu|(B)=0\), where \(B=\{x \in X: |h(x)| \neq 1\}\).
If we redefine \(h\) on \(B\) so that \(h(x)=1\) on \(B\) we obtain the function with the desired properties.$$\tag*{$\blacksquare$}$$
Suppose \(\mu\) is a positive measure on \((X,\mathcal M)\), and \(g \in L^1(X, \mu)\) and \[\lambda(E) = \int_Eg\,d\mu\quad \text{ for }\quad E \in \mathcal M.\] Then \(|\lambda|(E) =\int_E |g|\,d\mu\).
Proof. By the previous theorem there is \(h:X \to \mathbb{C}\), such that \(|h(x)|=1\) and \(d\lambda=h\,d|\lambda|,\) thus \[g\,d\mu=d\lambda= hd\,|\lambda|\] this gives \[d|\lambda|=\overline{h}g\,d\mu.\] Since \(|\lambda| \geq 0\) and \(\mu \geq 0\) it follows that \(\overline{h} g\geq 0\) \(\mu\)-a.e. so that \(\overline{h} g=|g|\), \(\mu\)-a.e., and this completes the proof.$$\tag*{$\blacksquare$}$$
If \(\nu\) is a complex measure on \((X,\mathcal M)\) then \[\bigg|\int_X f\,d\nu\bigg| \leq \int_X |f|\,d|\nu|\quad \text{ for any }\quad f \in L^1(X, \nu).\] Moreover, \(|\nu|(E)=\alpha(E)\), where \[\alpha(E)=\sup\bigg\{\bigg|\int_E f\,d\nu\bigg|: |f|\le 1\bigg\} \quad \text{ for any }\quad E\in \mathcal M.\]
Proof. By the polar decomposition \(d\nu=hd|\nu|,\) where \(|h|=1\), so \[\bigg|\int_X f\,d\nu\bigg|=\bigg|\int_X h f\,d|\nu|\bigg| \leq \int_X |f|\,d|\nu|.\] This shows \(\alpha(E)\le |\nu|(E)\). We also have \(\alpha(E)\ge \big|\int_E \overline{h}\,d\nu\big|=|\nu|(E)\). $$\tag*{$\blacksquare$}$$
Let \(\nu\) be a signed measure on \((X,\mathcal M)\), then there exist \(P, N\in \mathcal M\) such that \(P \cup N=X\) and \(P \cap N=\varnothing\) and \[\nu^+(E)=\nu(P\cap E) \quad \text{ and } \quad \nu^-(E)=-\nu(N\cap E) \quad \text{ for } \quad E\in \mathcal M.\] If \(P',N'\) is another such pair, then \(P \triangle P'=N \triangle N'\) is null for \(\nu\).
Proof. By the polar decomposition \(d\nu=hd|\nu|,\) where \(|h|=1\). But \(\nu\) is real, it follows that \(h\) is real (a.e., and therefore everywhere, by redefining on a set of measure \(0\)), hence \(h(x)=\pm 1\) for all \(x\in X\). Let \[P=\{x\in X: h(x)=1\} \quad \text{ and } \quad N=\{x\in X: h(x)=-1\}.\] Then \(P\cup N=X\) and \(P\cap N=\varnothing\).
Since \(\nu^+=\frac{1}{2}(|\nu|+\nu)\) and \[\frac{1}{2}(h(x)+1)= \begin{cases} h(x) & \text{ if } x\in P,\\ 0& \text{ if } x\in N, \end{cases}\] we obtain that \[\nu^+(E)=\frac{1}{2}\int_{E}(1+h)d|\nu|=\int_{E\cap P}hd|\nu|=\nu(P\cap E).\]
Since \(\nu(E)=\nu(P\cap E)+\nu(N\cap E)\) and \(\nu=\nu^+-\nu^-\) we obtain that \[\nu^+(E)=\nu^+(E)-\nu(E)=-\nu(N\cap E).\]
If \(P',N'\) is another pair of sets in the statement of theorem we have \(P \setminus P' \subseteq P\) and \(P \setminus P' \subseteq N'\), so \(P \setminus P'\) is both positive and negative hence null, likewise \(P' \setminus P\). Then \(P \triangle P'=N \triangle N'\) is null for \(\nu\).$$\tag*{$\blacksquare$}$$
Let \(\nu\) be a signed measure on \((X,\mathcal M)\), then there exist unique positive measures \(\nu^+\) and \(\nu^-\) such that \[\nu=\nu^+-\nu^- \quad \text{ and } \quad \nu^+ \perp \nu^-.\] Moreover, \(\nu^+,\nu^-\) are minimal in the sense that if \(\nu=\mu_1-\mu_2\) for two positive measures \(\mu_1, \mu_2\) on \((X,\mathcal M)\), then \(\mu_1\ge \nu^+\) and \(\mu_2\ge\nu^-\).
Proof. Let \(X=P \cup N\) be the Hahn decomposition for \(\nu\), then \[\nu^+(E)=\nu(P\cap E) \quad \text{ and } \quad \nu^-(E)=-\nu(N\cap E) \quad \text{ for } \quad E\in \mathcal M.\] Clearly \(\nu^+ \perp \nu^-\), since \(P \cap N=\varnothing\). Since \(\nu\le \mu_1\) and \(\nu\ge-\mu_2\) we obtain \[\begin{gathered} \nu^+(E)=\nu(P\cap E)\le \mu_1(P\cap E)\le \mu_1(E)\\ -\nu^-(E)=\nu(N\cap E)\ge -\mu_2(N\cap E)\ge -\mu_2(E). \end{gathered}\]
Finally we show uniqueness in the Jordan decomposition.
If \(\nu=\mu^+-\mu^-\) and \(\mu^+ \perp \mu^-\) let \(E,F \in \mathcal M\) be such that \(E \cap F=\varnothing\), \(E \cup F=X\) and \[\mu^+(F)=\mu^{-}(E)=0.\]
Then \(X=E \cup F\) is another Hahn decomposition for \(\nu\) so \(P \triangle E\) is \(\nu\)-null. Thus for any \(A \in \mathcal M\) we obtain \[\mu^+(A)=\mu^+(A \cap E)=\nu(A \cap E)=\nu(A \cap P)=\nu^+(A)\] and likewise \(\nu^-=\mu^-\).$$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a fixed measure space. Some observations are in order.
Suppose that \(p\) and \(q\) are conjugate exponents, i.e., \(1/p + 1/q = 1\). For each \(g \in L^q(X)\), we define a linear functional \(\phi_g\) on \(L^p(X)\) by \[\phi_g(f) = \int_X fgd\mu.\]
By Hölder’s inequality, we have \[|\phi_g(f)| = \Big| \int_X fgd\mu \Big| \leq \|g\|_{L^q} \|f\|_{L^p},\] so \(\phi_g\) is a bounded linear functional and \(\|\phi_g\|_{(L^p)^*}\le \|g\|_{L^q}\).
Especially in the \(p=2\) case, it may be more appropriate to define \(\phi_g(f) = \int_X f\overline{g}d\mu\), and the same convention can be used for \(p \neq 2\) without changing the results ahead in any essential way.
Recall that a measure \(\mu\) is semifinite if for each \(E\in \mathcal M\) with \(\mu(E)=\infty\) there exists \(F\in \mathcal M\) such that \(F\subseteq E\) and \(0<\mu(F)<\infty\).
Proposition. Let \((X, \mathcal M, \mu)\) be measure space. Suppose that the exponents \(p, q>0\) satisfy \(1 \leq q < \infty\) and \(1/p + 1/q = 1\). If \(g \in L^q\), then \[\|g\|_{L^q} = \|\phi_g\|_{(L^p)^*} = \sup\bigg\{\Big| \int_X fgd\mu \Big|: \|f\|_{L^p} = 1 \bigg\}.\] If \(\mu\) is semifinite, this result also holds for \(q = \infty\).
Proof. Hölder’s inequality gives that \(\|\phi_g\|_{(L^p)^*} \leq \|g\|_{L^q}\), and equality is trivial if \(g = 0\) almost everywhere. Otherwise, we show that the requisite supremum is attained for a specific function \(f\).
If \(1 < q < \infty\), let \[f = \frac{|g|^{q-1} \overline{{\rm sgn}{g}}}{\|g\|_{L^q}^{q-1}}.\]
Then \[\|f\|_{L^p}^{p} = \frac{\int_X |g|^{(q-1)p}d\mu}{\|g\|_{L^q}^{(q-1)p}} = \frac{\int_X |g|^qd\mu}{\int_X |g|^qd\mu} = 1,\] so \[\|\phi_g\|_{(L^p)^*} \geq \int_X fgd\mu = \frac{\int_X |g|^qd\mu}{\|g\|_{L^q}^{q-1}} = \|g\|_{L^q}.\]
If \(q = 1\), let \(f = \overline{{\rm sgn}{g}}.\) Then \(\|f\|_{L^\infty} = 1\) and \(\int_X fg d\mu= \|g\|_{L^1}\).
If \(q = \infty\), let \(\varepsilon > 0\) and let \[A = \{x\in X: |g(x)| > \|g\|_{L^\infty} - \varepsilon \}.\] Then \(\mu(A) > 0\), so, if \(\mu\) is semifinite, there exists \(B \subset A\) with \(0 < \mu(B) < \infty\). Let \(f = \mu(B)^{-1} \mathbf{1}_{{B}} \overline{{\rm sgn}{g}}.\) Then \(\|f\|_{L^1} = 1\) and \[\|\phi_g\|_{(L^1)^*} \geq \int_X fgd\mu = \frac{1}{\mu(B)} \int_B |g|d\mu \geq \|g\|_{L^\infty} - \varepsilon.\] Since \(\varepsilon > 0\) was arbitrary, this shows that \(\|\phi_g\|_{(L^1)^*} = \|g\|_{L^\infty}\).$$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be measure space. Let \(1 \leq p, q \le \infty\) and \(1/p + 1/q = 1\). Let \(g\) be a measurable function on \(X\) such that \(fg \in L^1(X)\) for all \(f\) in the space \(\Sigma\) of simple functions that vanish outside of a set of finite measure. Suppose that the quantity \[M_q(g) = \sup\bigg\{\Big| \int_X fgd\mu \Big|: f \in \Sigma \ \text{ and } \ \|f\|_{L^p}=1 \bigg\}\] is finite and suppose also that either the set \(S_g = \{x\in X: g(x) \neq 0\}\) is \(\sigma\)-finite or \(\mu\) is semifinite. Then \(g \in L^q(X)\) and \(M_q(g) = \|g\|_{L^q}\).
Proof. First, we remark that, if \(f\) is a bounded measurable function that vanishes outside a set \(E\) of finite measure and \(\|f\|_{L^p}=1\), then \[\Big|\int_Xfg d\mu\Big| \leq M_q(g).\]
Indeed, take a sequence \((f_n)_{n\in\mathbb N}\) of simple functions with \(|f_n| \leq |f|\) and \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) \(\mu\)-a.e. Since \(|f_n| \leq \|f\|_{L^\infty} \mathbf{1}_{{E}}\) and \(\mathbf{1}_{{E}}g \in L^1(X)\), by the (DCT), we have that \[\Big|\int_Xfg d\mu\Big|=\lim_{n\to \infty}\Big|\int_Xf_ng d\mu\Big|\leq M_q(g).\]
Now suppose that \(q < \infty\). We may assume that \(S_g\) is \(\sigma\)-finite since this condition automatically holds when \(\mu\) is semifinite. Show this!
Let \((E_n)_{n\in \mathbb N}\) be an increasing sequence of sets of finite measure such that \(S_g = \bigcup_{n\in \mathbb N} E_n\). Let \((\phi_n)_{n\in \mathbb N}\) be a sequence of simple functions such that \(\phi_n \ _{\overrightarrow{n\to \infty}} \ g\) pointwise and \(|\phi_n| \leq |g|\). Let \(g_n = \phi_n \mathbf{1}_{{E_n}}\). Then \(g_n \ _{\overrightarrow{n\to \infty}} \ g\) pointwise, \(|g_n| \leq |g|\), and \(g_n\) vanishes outside \(E_n\). Let \[f_n = \frac{|g_n|^{q-1} \overline{{\rm sgn}{g}}}{\|g_n\|_{L^q}^{q-1}}.\] We have \(\|f_n\|_{L^p} = 1\) as in the proof of the previous proposition.
By Fatou’s lemma, we have \[\begin{aligned} \|g\|_{L^q} &\leq \liminf_{n\to\infty} \|g_n\|_{L^q} = \liminf_{n\to\infty} \int_X |f_n g_n|d\mu \\ &\leq \liminf_{n\to\infty} \int_X |f_n g|d\mu = \liminf_{n\to\infty} \int_X f_n gd\mu \leq M_q(g). \end{aligned}\]
On the other hand, Hölder’s inequality gives \(M_q(g) \leq \|g\|_{L^q}\), so the proof is complete in the \(1\le q < \infty\) case.
If \(q = \infty\), let \(\varepsilon > 0\) and define \(A = \{x\in X: |g(x)| \geq M_\infty(g) + \varepsilon\}\). If \(\mu(A)>0\), we could choose \(B \subset A\) with \(0 < \mu(B) < \infty\) (using either that \(\mu\) is semifinite if \(\mu(A)=\infty\) or that \(A \subset S_g\)).
Setting \(f = \mu(B)^{-1} \mathbf{1}_{{B}} \overline{{\rm sgn}{g}}\), we would then have \(\|f\|_{L^1}=1\) and \(\int_X fg d\mu = \mu(B)^{-1} \int_B|g|d\mu \geq M_\infty(g) + \varepsilon\). But this contradicts our remark at the beginning of the proof. Therefore, \(\|g\|_{L^\infty} \leq M_\infty(g)\). The reverse inequality is obvious. $$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be measure space. Let \(1 \leq p, q \le \infty\) and \(1/p + 1/q = 1\). If \(1 < p < \infty\), then, for each \(\phi \in (L^p(X))^*\), there exists \(g \in L^q(X)\) such that \[\phi(f) = \int_X fgd\mu \quad \text{ for all } \quad f \in L^p(X).\] Hence, \(L^q(X)\) is isometrically isomorphic to \((L^p(X))^*\). The same conclusion holds for \(p=1\) provided that \(\mu\) is \(\sigma\)-finite.
Proof. First, suppose that \(\mu(X)<\infty\), so that all simple functions are in \(L^p(X)\). If \(\phi \in (L^p(X))^*\) and \(E\in \mathcal M\), let \(\nu(E) = \phi(\mathbf{1}_{{E}})\). For any disjoint sequence \((E_j)_{j\in \mathbb N}\subseteq \mathcal M\), if \(E = \bigcup_{j\in \mathbb N} E_j\), we have \(\mathbf{1}_{{E}} = \sum_{j\in \mathbb N} \mathbf{1}_{{E_j}}\), where the series converges in \(L^p(X)\) norm: \[\Big\| \mathbf{1}_{{E}} - \sum_{j\in \mathbb N} \mathbf{1}_{{E_j}} \Big\|_{L^p} = \Big\|\sum_{j=n+1}^\infty \mathbf{1}_{{E_j}} \Big\|_{L^p} = \mu\Big(\bigcup_{j=n+1}^\infty E_j\Big)^{1/p} \ _{\overrightarrow{n\to \infty}} \ 0.\]
(Note that this step is where we require that \(p < \infty\)). Hence, since \(\phi\) is linear and continuous, we have \[\nu(E)= \phi(\mathbf{1}_{{E}}) = \sum_{j\in \mathbb N} \phi(\mathbf{1}_{{E_j}}) = \sum_{j\in \mathbb N} \nu(E_j),\] so \(\nu\) is a complex measure. Also, if \(\mu(E) = 0\), then \(\mathbf{1}_{{E}} = 0\) considered as an element of \(L^p(X)\), so \(\nu(E) = 0\). This shows that \(\nu \ll \mu\).
By the Radon–Nikodym theorem, there exists \(g \in L^1(X)\) such that \[\phi(\mathbf{1}_{{E}}) = \nu(E) = \int_E g d\mu \quad \text{ for all } \quad E\in\mathcal M.\]
Therefore, \(\phi(f) = \int_X fg d\mu\) for all simple functions \(f\). Moreover, \(\big| \int_X fgd\mu \big| \leq \|\phi\|_{(L^p)^*} \|f\|_{L^p}\), so \(g \in L^q(X)\) by the previous theorem.
Since simple functions are dense in \(L^p(X)\), we have \(\phi(f) = \int_X fgd\mu\) for all \(f \in L^p(X)\).
Now suppose that \(\mu\) is \(\sigma\)-finite. Let \((E_n)_{n\in\mathbb N}\) be an increasing sequence of sets such that \(0 < \mu(E_n) < \infty\) and \(X = \bigcup_{n\in\mathbb N} E_n\).
We identify \(L^p(E_n)\) and \(L^q(E_n)\) with the subspaces of \(L^p(X)\) and \(L^q(X)\) respectively consisting of functions that vanish outside \(E_n\).
The preceding argument shows that, for each \(n\), there exists \(g_n \in L^q(E_n)\) such that \(\phi(f) = \int_X fg_nd\mu\) for all \(f \in L^p(E_n)\) and \(\|g_n\|_{L^q(E_n)} = \| \phi\|_{(L^p(E_n))^*} \leq \|\phi\|_{(L^p)^*}\).
The function \(g_n\) is unique up to a nullset, so \(g_n = g_m\) almost everywhere on \(E_n\) for \(n < m\). Therefore, we can define \(g\) almost everywhere on \(X\) by setting \(g = g_n\) on \(E_n\).
Moreover, if \(f \in L^p(X)\), then, by the (DCT), we have \(f \mathbf{1}_{{E_n}} \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^p(X)\) norm, and consequently \[\phi(f) = \lim_{n\to \infty}\phi(f \mathbf{1}_{{E_n}}) = \lim_{n\to \infty} \int_{E_n} fgd\mu = \int_X fg d\mu.\]
Finally, suppose that \(\mu\) is arbitrary and \(p > 1\), so \(q < \infty\). For each \(\sigma\)-finite set \(E \in \mathcal M\), there is a unique \(g_E \in L^q(E)\) (up to a nullset) such that \(\phi(f) = \int_X f g_Ed\mu\) for all \(f \in L^p(E)\) and \(\|g_E\|_{L^q} \leq \|\phi\|_{(L^p)^*}\).
If \(F\) is \(\sigma\)-finite and \(F \supset E\), then \(g_F = g_E\) \(\mu\)-a.e. on \(E\), hence we have \(\|g_F\|_{L^q} \geq \|g_E\|_{L^q}\). Let \(M\) be the supremum of \(\|g_E\|_{L^q}\) as \(E\) ranges over all \(\sigma\)-finite sets, noting that \(M \leq \|\phi\|_{(L^p)^*}\). Choose a sequence \((E_n)_{n\in \mathbb N}\) so that \(\|g_{E_n}\|_{L^q} \ _{\overrightarrow{n\to \infty}} \ M\) and set \(F = \bigcup_{n\in \mathbb N} E_n\). Then \(F\) is \(\sigma\)-finite and \(\|g_F\|_{L^q} \geq \|g_{E_n}\|_{L^q}\) for all \(n\in \mathbb N\). Therefore, \(\|g_F\|_{L^q} = M\).
Now, if \(A\) is a \(\sigma\)-finite set containing \(F\), we have \[\int_X |g_F|^qd\mu + \int_X |g_{A \setminus F}|^qd\mu = \int_X |g_A|^qd\mu \leq M^q = \int_X |g_F|^qd\mu.\]
Thus, \(g_{A \setminus F} = 0\) and \(g_A = g_F\) almost everywhere. (Here we used that \(q < \infty\)). But if \(f \in L^p(X)\), then \(A = F \cup \{x\in X: f(x) \neq 0\}\) is \(\sigma\)-finite, so \(\phi(f) = \int_X fg_Ad\mu = \int_X fg_Fd\mu\). Thus, we may take \(g = g_F\), and the proof is complete. $$\tag*{$\blacksquare$}$$