\( \newcommand{\nat}{\mathbb{N}} \newcommand{\real}{\mathbb{R}} \newcommand{\borel}{\mathbb{B}} \newcommand{\sM}{\mathcal{M}} \newcommand{\comp}{\mathbb{C}} \newcommand{\qed}{\mbox{\rule{.4em}{1.7ex}\hspace{.6em}}} \)

\(L^p\) spaces

Muruhan Rathinam

January 1, 2026

1 Brief review of measure theory

We assume familiarity with measure spaces and the Lebesgue integral. We briefly recall that given a nonempty set \(S\), a subset \(\Sigma \subset 2^S\) of the power set of \(S\) is said to be a \(\sigma \)-algebra provided \(\emptyset \in \Sigma \), for each \(A \in \Sigma \), \(A^c=S \setminus A \in \Sigma \) and for each countable collection \(A_n\) (\(n \in \nat \)) with \(A_n \in \Sigma \) for all \(n \in \nat \) it follows that \(\bigcup _n A_n \in \Sigma \).

Given a nonempty set \(S\), the power set \(2^S\) is the largest \(\sigma \)-algebra on \(S\) and \(\{\emptyset ,S\}\) is the smallest \(\sigma \)-algebra. The reason for not always working with the power set \(2^S\) is that when \(S\) is an uncountable set it turns out that it may not be possible to define a “size” (measure) for every subset of \(S\) in a desirable manner. A critical example is \(S=\real \). It is well known that it is not possible to define a “length” for every subset \(A \subset \real \) that extends the notion of length of bounded intervals of \(\real \) in a meaningful way. It turns out that we can define a length for a large class of subsets of \(\real \) in a meaningful way. One example of this large class is the Borel \(\sigma \)-algebra \(\borel \) which is defined to be the intersection of all \(\sigma \)-algebras on \(\real \) that contain the collection of all open sets.

A measure \(\mu \) on \((S,\Sigma )\) is a map \(\mu :\Sigma \to [0,\infty ]\) that satisfies \(\mu (\emptyset )=0\) and

\[ \mu \left (\bigcup _n A_n\right ) = \sum _n \mu (A_n) \]
for all sequences \(A_n \in \Sigma \) with \(A_n \cap A_m = \emptyset \) whenever \(n \neq m\).

Let \((S,\Sigma ,\mu )\) be a measure space. A function \(f:S \to \real \) is said to be measurable provided \(f^{-1}(E) \in \Sigma \) for each Borel set \(E \in \borel \). The theory of Lebesgue integration provides for a way to define the integral

\[ \int _A f \, d\mu \]
for every nonnegative and measurable function \(f:S \to [0,\infty )\) where \(A \in \Sigma \). The above integral may be \(\infty \). Given a function \(f:S \to \real \) we define the positive and negative parts \(f^+\) and \(f^-\) of \(f\) by
\[ f^+(x) = \max \{f(x),0\}, \quad f^-(x)=\max \{-f(x),0\}. \]
We note that \(f=f^+-f^-\) and \(|f|=f^+ + f^-\). It turns out that if \(f:S \to \real \) is measurable, then so are \(f^+\) and \(f^-\). If
\[ \int _S |f| \, d\mu = \int _S (f^+ + f^-) \, d\mu < \infty , \]
then we say that \(f\) is integrable and define
\[ \int _A f\, d\mu = \int _A f^+ \, d\mu - \int _A f^- \, d\mu . \]
Note that we have avoided \(\infty - \infty \).

Two measurable functions \(f,g:S \to \real \) are said to be equal almost everywhere provided

\[ \mu \{x \in S \, | \, f(x) \neq g(x)\} =0. \]
We write this as
\[ f=g \; \text {a.e.} \]
If \(f\) is integrable and \(f=g\) a.e. where \(g\) is also measurable, then
\[ \int _A f \, d\mu = \int _A g \, d\mu , \]
for all \(A \in \Sigma \). From the point of view of measure theory, the two functions \(f\) and \(g\) are the same, and hence we could “identify” them. This creates an equivalence relation.

2 \(L^p\) spaces

For \(1 \leq p <\infty \) the space \(L^p(S,\Sigma ,\mu )\) is defined to be the set of equivalence classes of measurable functions \([f]\) such that

\[ \int _S |f|^p \, d\mu < \infty . \]
Here \([f]\) denotes the equivalence class of \(f:S \to \real \).

A measurable function \(f:S \to \real \) is said to be essentially bounded provided there exists a bounded function \(g:S \to \real \) (there exists \(M>0\) such that \(|g(x)|\leq M\) for all \(x \in S\)) such that \(f=g\) a.e. Given an essentially bounded measurable function \(f\) we define the essential supremum of \(f\) by

\[ \text {ess sup} f = \inf \{M>0 \, | \, f \leq M \, \text {a.e.}\}. \]
We note that if \(f=g\) a.e. where \(f\) and \(g\) are essentially bounded measurable functions then their essential suprema are equal. The space \(L^\infty (S,\Sigma ,\mu )\) is defined to be the set of equivalence classes of functions that are essentially bounded.

When the measure space \((S,\Sigma ,\mu )\) is clear from the context, we simply write \(L^p\) in place of \(L^p(S,\Sigma ,\mu )\). Having defined the \(L^p\) for \(1 \leq p \leq \infty \), we first note that \(L^p\) is a vector space over \(\real \). This is easy to establish and left as an exercise and we observe that the inequality

\[ (a+b)^p \leq 2^p (a^p + b^p) \quad \forall a,b,p \geq 0, \]
is useful.

It turns out that we can define a norm on \(L^p\) as follows. If \(f \in L^p\) for some \(p \in [1,\infty )\) then

\[ \|f\|_p = \left ( \int _S |f|^p \, d\mu \right )^{\frac {1}{p}} \]
defines a norm on \(L^p\). If \(f \in L^\infty \)
\[ \|f\|_\infty = \text {ess sup} |f| \]
defines a norm on \(L^\infty \).

Recall that a norm \(\|.\|\) on a vector space \(V\) over \(\real \) satisfies nonnegativity (\(\|v\| \geq 0\)), positive definiteness (\(\|v\|=0\) if and only if \(v=0\)), homogeneity (\(\|\alpha v\|=|\alpha |\|v\|\)) and the triangle inequality (\(\|u+v\| \leq \|u\| + \|v\|\)). Nonnegativity and homogeneity are easy to establish. Positive definiteness is established via measure theoretic considerations. In the case of \(p=1\), the triangle inequality follows from triangle inequality of real numbers and monotonicity of Lebesgue integrals (\(f \leq g\) implies \(\int _S f d\mu \leq \int _S g d\mu \)). In the case of \(p=\infty \), the triangle inequality follows from the triangle inequality for real numbers and properties of suprema. Establishing the triangle inequality for the case of \(1<p<\infty \) is more involved. In fact it is known as the  em Minkowski inequality and we shall establish it by first proving certain important inequalities.

Conjugates: Let \(1 < p < \infty \). Then the conjugate of \(p\) is the unique number \(1<q<\infty \) that satisfies \(1/p+1/q=1\). Moreover, the conjugate of \(1\) is defined to be \(\infty \) and that of \(\infty \) is defined to be \(1\). Thus being conjugates is a symmetric relationship.

3 Important inequalities

Young’s inequality: Let \(1<p<\infty \) and \(q\) be the conjugate of \(p\). Then for \(a,b \geq 0\)

\[ ab \leq \frac {a^p}{p} + \frac {b^q}{q}. \]

Before we state the Holder’s inequality, we define the function \(\text {sgn}:\real \to \real \) by \(\text {sgn}(x)=1\) for \(x \geq 0\) and \(\text {sgn}(x)=-1\) for \(x<0\).

Theorem 1 (Holder’s inequality) Let \(1 \leq p <\infty \) and \(q\) be the conjugate of \(p\). Let \(f \in L^p(S,\Sigma ,\mu )\) and \(g \in L^q(S,\Sigma ,\mu )\). Then \(fg\) is integrable and

\[ \int _S |fg| \, d\mu \leq \|f\|_p \|g\|_q. \]
If \(f \neq 0\), the function \(f^*:S \to \real \) defined by
\[ f^*(x) = \|f\|_p^{1-p} \, \text {sgn}(f(x)) \, |f(x)|^{p-1} \]
belongs to \(L^q\),
\[ \int _S f \, f^* \, d\mu = \|f\|_p, \]
and \(\|f^*\|_q=1\).

Proof  First we consider the case of \(p=1\). Then \(q=\infty \). If \(f \in L^1\) and \(g \in L^\infty \) then \(|g| \leq \|g\|_\infty \) a.e. Thus

\[ \int _S |fg| \, d\mu \leq \|g\|_\infty \int _S |f| \, d\mu = \|g\|_\infty \|f\|_1. \]
Moreover \(f^*= \text {sgn}(f)\), so \(f \, f^*=|f|\). Thus
\[ \int _S f \, f^* \, d\mu = \int _S |f| \, d\mu = \|f\|_1 \]
and since \(|f^*|=1\), it also follows that \(\|f^*\|_\infty =1\). This proves the \(p=1\) case.

Let \(1<p<\infty \). Then if \(f=0\) or \(g=0\) the inequality holds trivially. Thus it suffices to suppose \(f \neq 0\) and \(g \neq 0\). We first consider the case \(\|f\|_p=\|g\|_q=1\). By Young’s inequality

\[ |fg| \leq \frac {|f|^p}{p} + \frac {|g|^q}{q}. \]
By monotonicity of the integral
\[ \int _S |fg| \, d\mu \leq \int _S \left ( \frac {|f|^p}{p} + \frac {|g|^q}{q}\right ) \, d\mu = \frac {\|f\|_p^p}{p} + \frac {\|g\|^q_q}{q} = \frac {1}{p} + \frac {1}{q} = 1. \]
Thus the Holder’s inequality is proven for \(\|f\|_p=\|g\|_q=1\) case. For general \(f \neq 0\) and \(g \neq 0\) we apply the above result to \(f/\|f\|_p\) and \(g/\|g\|_q\) to obtain
\[ \frac {\int _S |fg| \, d\mu }{\|f\|_p \|g\|_q} \leq 1. \]
Multiplying by \(\|f\|_p \|g\|_q\) implies the result.

Now, from the definition of \(f^*\), it follows that

\[ f \, f^* = \|f\|_p^{1-p} \, |f|^p, \]
and hence
\[ \int _S f \, f^* \, d\mu = \|f\|_p^{(1-p)+p} = \|f\|_p. \]
Moreover, since \(|f^*|=\|f\|_p^{1-p} |f|^{p-1}\), we have that
\[ |f^*|^q = \|f\|_p^{q(1-p)} \, |f|^{q(p-1)}. \]
Noting that \(q(p-1)=p\), we conclude that
\[ \int _S |f^*|^q \, d\mu = \|f\|_p^{-p} \|f\|_p^p = 1. \]
Thus \(\|f^*\|_q =1\), completing the proof. _

We shall refer to the function \(f^*\) corresponding \(f \neq 0\) defined above as the conjugate function of \(f\).

Corollary 2 If \(f,g \in L^2\) then

\[ \int _S |f g| \, d\mu \leq \sqrt {\int _S |f|^2 \, d\mu } \, \sqrt {\int _S |g|^2 \, d\mu } \]

Theorem 3 (Minkowski inequality) Let \(f, g \in L^p\) where \(1 \leq p \leq \infty \). Then

\[ \|f+g\|_p \leq \|f\|_p + \|g\|_p. \]

Proof  The case of \(p=1\) and \(p=\infty \) are straightforward as mentioned earlier. This we consider \(1<p<\infty \). If \(f+g=0\) the result is trivial. So we consider the case of \(f+g \neq 0\). Then by the second part of the theorem on Holder’s inequality,

\[ \|f+g\|_p = \int _S (f+g) \, (f+g)^* \, d\mu = \int _S f (f+g)^* \, d\mu + \int _S g (f+g)^* \, d\mu , \]
where \((f+g)^*\) is the conjugate function of \(f+g\). Applying (the first and then the second parts of) the Holder’s inequality, we obtain
\[ \|f+g\|_p \leq \|f\|_p \|(f+g)^*\|_q + \|g\|_p \|(f+g)^*\|_q = \|f\|_p + \|g\|_q. \]
_

With the Minkowski inequality, it is clear that \(L^p\) is a normed vector space for \(1 \leq p \leq \infty \). Finally, we note that \(L^p\) is indeed a Banach space. In other words it is a complete normed vector space. The theorem that asserts this is referred to as the Riesz-Fischer Theorem by some authors including Royden.

4 \(L^p\) spaces are complete

Theorem 4 The normed vector space \(L^p(S,\Sigma ,\mu )\) is complete and hence a Banach space.

Proof  We show the proof for \(1 \leq p < \infty \). Let \(f_n \in L^p\) for \(n \in \nat \) and suppose that

\[ \sum _{n \in \nat } \|f_n\|_p < \infty . \]
Let \(g_n=\sum _{j=1}^n f_j\) be the sequence of partial sums. If we show that there exists \(g \in L^p\) such that \(g_n \to g\) in \(L^p\) we are done.

Let \(M>0\) satisfy \(\sum _{n \in \nat } \|f_n\|_p \leq M\). Define the sequence \(h_n:S \to \real \) by \(h_n = \sum _{j=1}^n |f_j|\). Since \((h_n)\) is an increasing sequence it has a limit (in the almost everywhere sense) which may be infinity. Let \(h = \lim _{n \to \infty } h_n \leq \infty \) a.e. Thus \(h_n^p \uparrow h^p\) a.e. as \(n \to \infty \). By monotone convergence theorem

\[ \int _S h_n^p \, d\mu \to \int _S h^p \, d\mu . \]
By Minkowski inequality, for each \(n\)
\[ \|h_n\|_p \leq \sum _{j=1}^n \|f_j\|_p. \]
Note that \(\| |f_j| \|_p=\|f_j\|_p\). Hence \(\|h_n\|_p \leq M\) for all \(n\). Thus for all \(n\)
\[ \int _S h_n^p \, d\mu \leq M^p \]
and hence \(\int _S h^p d\mu \leq M^p\). Therefore \(h< \infty \) a.e. Thus for almost all \(x \in S\), the series
\[ \sum _{n \in \nat } f_n(x) \]
converges absolutely and hence converges. Let \(g=\lim _{n \to \infty }g_n\) a.e. It follows that \(|g_n-g|^p \to 0\) a.e. as \(n \to \infty \). Since
\[ |g_n-g|^p \leq 2^p \left ( |g_n|^p + |g|^p \right ) \leq 2^p \left ( h_n^p + h^p \right ) \leq 2^{p+1} h^p \]
and \(h^p\) is integrable, by the dominated convergence theorem \(\int _S |g_n-g|^p d\mu \) converges to \(0\). That is \(g_n \to g\) in \(L^p\). _

5 Some examples of \(L^p\) spaces

We consider special examples of \(L^p(S,\Sigma ,\mu )\). First consider the case of a finite set \(S\) which we may without loss of generality take to be \(S=\{1,2,\dots ,n\}\). Take \(\Sigma = 2^S\) (the power set) and the measure \(\mu \) to be the counting measure. This means \(\mu (A)\) equals the number of elements in \(A\). Then a function \(f \in L^p(S,\Sigma ,\mu )\) is simply any function from \(S\) into \(\real \) and hence can be identified with \(n\)-tuples of real numbers. Thus \(L^p(S,\Sigma ,\mu )\) is the same as \(\real ^n\) and

\[ \|f\|_p = \left (\sum _{i=1}^n |f(i)|^p\right )^\frac {1}{p} \]
the \(p\)-norm of a vector in \(\real ^n\) (for \(p<\infty \)) and for \(p=\infty \) it is simply the sup-norm. We note that the integral with respect to the counting measure becomes a sum over the finite number (\(n\)) of elements.

Another example is where \(S=\nat \), \(\Sigma =2^\nat \) (the power set) and \(\mu \) the counting measure. In this case \(L^p(S,\Sigma ,\mu )\) becomes \(\ell ^p\) discussed earlier. The integral becomes an infinite sum.

In the study of PDEs, it is common to consider \(S \subset \real ^n\) to be a subset with nonempty interior, \(\Sigma =\borel (S)\), that is the \(\sigma \)-algebra generated by open subsets of \(S\) and \(\mu \) to be the \(n\)-dimensional Lebesgue measure.

6 Simple functions

Given a measure space \((S,\Sigma ,\mu )\) we recall the concept of a simple measurable function or just a simple function. A measurable function \(f:S \to \real \) is said to be simple provided \(f(S)\) is a finite set. Given a simple function \(f:S \to \real \) it follows that if we denote \(f(S)=\{a_1,\dots ,a_n\}\) and let \(A_i = f^{-1}(\{a_i\}) \in \Sigma \) then

\[ f = \sum _{i=1}^n a_i 1_{A_i} \]
where \(1_{A_i}\) is the indicator function of the set \(A_i\). This means \(1_{A_i}(x)=1\) for \(x \in A_i\) and \(1_{A_i}(x)=0\) for \(x \notin A_i\). It is easy to see that the set of all simple functions is a vector space over \(\real \). We also note that all simple functions belong to \(L^p\) for \(1 \leq p \leq \infty \). We shall show that the simple functions are dense in \(L^p\) for \(1 \leq p \leq \infty \).

Lemma 5 Let \(f \in L^p(S,\Sigma ,\mu )\) where \(1 \leq p <\infty \). Then for every \(\epsilon >0\) there exists \(A \in \Sigma \) such that \(\mu (A)<\infty \) and \(\int _{S \setminus A} |f|^p d\mu <\epsilon \).

Proof  Let \(g=|f|\). For each \(n \in \nat \) define \(E_n=g^{-1}([1/n,\infty ))\). Then \(E_n \in \Sigma \) and \(E_n \uparrow E\) where \(E=\cup _n E_n\). It follows that \(g=0\) on \(S \setminus E\). We also have that \(1_{E_n} g \uparrow 1_E g\) and by monotone convergence theorem

\[ \int _{E_n} g \, d\mu = \int _S 1_{E_n} g \, d\mu \, \big \uparrow \int _S 1_E g \, d\mu = \int _E g \, d\mu = \int _S g \, d\mu . \]
Thus given \(\epsilon >0\) there exists \(n \in \nat \) such that
\[ \int _{S \setminus E_n} g \, d\mu =\int _S g \, d\mu - \int _{E_n} g \, d\mu <\epsilon . \]
Finally, since \(g \geq 1/n\) on \(E_n\),
\[ \int _S g^p \, d\mu \geq \int _{E_n} g^p \, d\mu \geq \mu (E_n)/n. \]
Hence \(\mu (E_n) \leq n \|f\|^p_p < \infty \). _

Lemma 6 The space of simple functions is a dense subspace of \(L^p\) for \(1 \leq p \leq \infty \).

Proof  We first prove the \(p<\infty \) case. Let \(f \in L^p\) and let \(\epsilon >0\). Then there exists \(A \in \Sigma \) such that \(\mu (A)<\infty \) and \(\int _{S \setminus A}|f|^p d\mu < (\epsilon /2)^p\). Let \(g = 1_A f\). Then \(\|g-f\|_p < \epsilon /2\).

We define the sequence \(f_n\) of simple functions by defining their positive and negative parts \(f_n^+\) and \(f_n^-\) as follows. For \(n \in \nat \)

\[ f^+_n(x) = j \, 2^{-n} \]
provided \(j 2^{-n} \leq g^+(x) < (j+1) 2^{-n}\) for some integer \(0 \leq j < n 2^n -1\),
\[ f_n^+(x)= n \]
otherwise (that is \(g^+(x) \geq n\)). Likewise define
\[ f^-_n(x) = j \, 2^{-n} \]
provided \(j 2^{-n} \leq g^-(x) < (j+1) 2^{-n}\) for some integer \(0 \leq j < n 2^n -1\),
\[ f_n^-(x)= n \]
otherwise (that is \(g^-(x) \geq n\)). Define \(f_n=f^+_n-f^-_n\).

We observe that for all \(n\), \(f_n=0\) on \(S \setminus A\). For \(n \in \nat \) define \(E_n= g^{-1}([-n,n]) \cap A\). Then \(|f_n-g| < 2^{-(n-1)}\) on \(E_n\) for all \(n\). Hence

\[ \int _{E_n} |f_n-g|^p \, d\mu \leq 2^{-(n-1) p} \mu (E_n). \]
For all \(n\), since \(E_n \subset A\), we have that \(\mu (E_n)<\infty \). Hence \(\int _{E_n} |f_n-g|^p \, d\mu \to 0\) as \(n \to \infty \).

Moreover, on \(A \setminus E_n\) we have that \(|f_n-g|=|g|-n < |g|\), and hence

\[ \int _{A \setminus E_n} |f_n-g|^p \, d\mu \leq \int _{A \setminus E_n} |g|^p \, d\mu . \]
We also observe that \(E_n \uparrow E\) where \(E = \cup _n E_n= A\) since we can suppose that for each \(x \in S\), \(g(x) \in \real \) (not \(+\infty \) or \(-\infty \)). Hence \(1_{A \setminus E_n} |f_n-g|^p \to 0\) a.e. Since \(1_{A \setminus E_n} |f_n-g|^p \leq |g|^p\) and \(|g|^p\) is integrable, by the dominated convergence theorem
\[ \int _{A \setminus E_n} |f_n-g|^p \, d\mu \to 0. \]
Thus, noting that \(f_n\) and \(g\) vanish off of \(A\), we have
\[ \int _S |f_n-g|^p \, d\mu = \int _{E_n} |f_n-g|^p \, d\mu + \int _{A \setminus E_n} |f_n-g|^p \, d\mu \to 0. \]
Consequently, there exists \(n \in \nat \) such that the simple function \(f_n\) satisfies \(\|f_n-g\|_p < \epsilon /2\). Hence \(\|f_n-f\|_p < \epsilon \), which completes the proof for \(p<\infty \) case.

For the \(p=\infty \) case, let \(f \in L^\infty \). Define a sequence \((f_n)\) of simple functions as before.

\[ f_n^\pm (x) = j 2^{-n} \]
if \(j 2^{-n} \leq f^\pm (x) < (j+1) 2^{-n}\) for some \(0 \leq j < n 2^n -1\). Otherwise \(f_n^\pm (x) = n\). Then \(|f_n-f|< 2^{-(n-1)}\) a.e. for all \(n > \|f\|_\infty \). Thus \(\|f_n-f\|_\infty \to 0\). This proves the \(p=\infty \) case. _

Corollary 7 Let \(\phi \in (L^p)^*\) be a continuous linear functional. Suppose \(\phi (1_A)=0\) for all \(A \in \Sigma \). Then \(\phi =0\).

7 Riesz representation theorem

In the study of \(\ell ^p\) spaces we showed that for \(1 \leq p <\infty \), the dual space of \(\ell ^p\) is isometrically isomorphic to \(\ell ^q\) where \(q\) is the conjugate of \(p\). This result is a special case of the more general result also known as the Riesz representation theorem.

Theorem 8 (Riesz Representation Theorem) Let \((S,\Sigma ,\mu )\) be a \(\sigma \)-finite measure space. We denote \(L^p(S,\Sigma ,\mu )\) by \(L^p\). Let \(1 \leq p < \infty \) and \(q\) be its conjugate. Define the operator \(R:L^q \to {(L^p)}^*\) by

\[ R(g)(f) = \int _S f g \, d\mu , \]
for all \(g \in L^q\) and \(f \in L^p\). Then \(R\) is an isometric isomorphism.

Proof  We sketch the proof here. First we note that that \(fg\) is integrable for all \(g \in L^q\) and all \(f \in L^p\) by the Holder’s inequality. Thus, \(R(g)\) defines a linear functional on \(L^p\). Also, from the Holder’s inequality, \(|R(g)(f))|=|\int _S fg d\mu | \leq \|f\|_p \|g\|_q\) and thus \(R(g)\) is a bounded linear functional. This ensures that \(R\) is well defined,. It is easy to see that \(R\) is linear, and it is also clear that \(\|R(g)\| \leq \|g\|_q\).

Next we show that \(\|R(g)\|=\|g\|_q\). If \(g=0\) this is obvious. If \(g \neq 0\), then we consider the case of \(1<p<\infty \) so that \(1<q<\infty \). Then, choosing \(f=g^*\) to be the conjugate function defined in the Holder’s inequality theorem, we see that \(f \in L_p\), \(\|f\|_p=1\) and that \(R(g)(f)=\int _S fg d\mu = \|g\|_q\). This shows that \(\|R(g)\| \geq \|g\|_q\). So we conclude that \(\|R(g)\|=\|g\|_q\) (for the case of \(1<p<\infty \)). If \(p=1\), then \(q=\infty \), and the conjugate function of \(g \in L^\infty \) is not defined. So we proceed differently. Let \(\epsilon >0\). Then the set \(G=\{x \in S \, | \, |g(x)| > \|g\|_\infty - \epsilon \}\) has positive measure. Since \(\mu \) is a \(\sigma \)-finite, there exists a set \(A \in \Sigma \) such that \(0<\mu (A)<\infty \) and \(A \subset G\). Define

\[f= \frac {\text {sgn}(g) \, 1_A}{\mu (A)}.\]
Then \(f \in L_1=L_p\), \(\|f\|_1=1\) and \(R(g)(f)=\int _S fg d \mu = \frac {1}{\mu (A)} \int _A |g| d\mu \geq \|g\|_\infty - \epsilon \). As this holds for all \(\epsilon >0\), we conclude that
\[ \sup \left \{ |R(g)(f)| \,\, | \, f \in L_1, \,\, \|f\|_1=1\right \} \geq \|g\|_\infty . \]
Hence \(\|R(g)\| \geq \|g\|_\infty \). Since the reverse inequality was shown earlier, we conclude that \(\|R(g)\|=\|g\|_\infty \).

Thus, we have established that \(R\) is a linear isometry from \(L^q\) into \((L^p)^*\). As a linear isometry, it is an injection. It remains to show that it is a surjection. Given \(\varphi \in (L^p)^*\) with \(\varphi \neq 0\), we need to show that there exists \(g \in L^q\) such that \(R(g)=\varphi \). This is done via the use of the Radon-Nikodym theorem.

We only provide a sketch of the proof. First one considers the case when \(\mu (S)<\infty \). Define \(\nu :\Sigma \to \real \) as follows. For each \(A \in \Sigma \), we set \(\nu (A) = \varphi (1_A)\). This makes sense, since \(1_A \in L^p\) as \(\mu (A)<\infty \). Then it can be shown that \(\nu \) is a finite signed measure and that \(\nu \ll \mu \). Then by the Radon-Nikodym theorem there exists an integrable function \(g\) such that \(\varphi (1_A)=\nu (A)=\int _A g d\mu = \int _S 1_A g d\mu \) for all \(A\). Then, one proves that \(\varphi (f)=\int _S f g d\mu \) for all \(f \in L^p\). Finally, one extends the result to \(\sigma \)-finite \(\mu \). At this juncture, one obtains \(g \in L^q\) such that \(\varphi (f)=\int _S f g d\mu = R(g)(f)\). _