The [Lebesgue integral](/page/Lebesgue%20Integral) solves a basic problem of approximation theory: if a sequence of functions converges in a reasonable sense, does the integral of the limit equal the limit of the integrals? The Riemann integral fails this test badly --- there exist sequences of Riemann-integrable functions, converging pointwise to a limit that is not Riemann integrable. The Lebesgue integral, built on [measure theory](/page/Hausdorff%20Measure), handles such limits gracefully through the [Monotone Convergence Theorem](/theorems/509) and the [Dominated Convergence Theorem](/theorems/4).

But a deeper problem remains. In applications to partial differential equations, to the [calculus of variations](/page/Calculus%20of%20Variations), and to probability theory, we do not merely need to integrate individual functions --- we need a **space** of functions that behaves like a well-structured infinite-dimensional vector space. Specifically, we need:

1. **Completeness.** [Cauchy sequences](/page/Cauchy%20Sequence) of functions must converge to a function *in the same space*. Without this, iterative methods (Picard iteration, Galerkin approximation, variational minimization) cannot be guaranteed to converge. 2. **A norm.** We need a quantitative measure of the "size" of a function that is compatible with the vector space structure. 3. **Duality.** We need a concrete description of the bounded linear functionals on the space, so that weak solutions of differential equations and distributional formulations have a precise meaning.

The classical space $C([0,1])$ of [continuous functions](/page/Continuity%20(Metric%20Spaces)), equipped with the $L^2$-type norm, fails the first requirement catastrophically.

[example: Failure of Completeness in $C([0,1])$] Let $U = (0,1) \subset \mathbb{R}$ and consider the space $C([0,1])$ equipped with the norm \begin{align*} \|f\|_2 := \left( \int_0^1 |f(t)|^2 \, d\mathcal{L}^1(t) \right)^{1/2}. \end{align*} Define the sequence of [continuous](/page/Continuity) functions $f_m: [0,1] \to \mathbb{R}$ by \begin{align*} f_m(t) := \begin{cases} 0 & \text{if } 0 \le t \le \tfrac{1}{2}, \\ m(t - \tfrac{1}{2}) & \text{if } \tfrac{1}{2} < t < \tfrac{1}{2} + \tfrac{1}{m}, \\ 1 & \text{if } \tfrac{1}{2} + \tfrac{1}{m} \le t \le 1. \end{cases} \end{align*} Each $f_m$ is continuous, so $f_m \in C([0,1])$. We verify that $\{f_m\}$ is Cauchy. For $m < k$, the functions $f_m$ and $f_k$ differ only on the interval $(\tfrac{1}{2}, \tfrac{1}{2} + \tfrac{1}{m})$, where both take values in $[0,1]$. Thus \begin{align*} \|f_m - f_k\|_2^2 = \int_{1/2}^{1/2 + 1/m} |f_m(t) - f_k(t)|^2 \, d\mathcal{L}^1(t) \le \int_{1/2}^{1/2 + 1/m} 1 \, d\mathcal{L}^1(t) = \frac{1}{m} \to 0. \end{align*} The pointwise limit is $f(t) = \mathbb{1}_{(1/2, 1]}(t)$, the indicator function of $(1/2, 1]$. This function is *not* continuous, so $f \notin C([0,1])$. The sequence is Cauchy but does not converge within the space. The norm $\|\cdot\|_2$ is perfectly well-defined on $C([0,1])$, but the resulting [normed space](/page/Normed%20Vector%20Space) is incomplete. [/example]

The resolution is to enlarge the space by passing from continuous functions to [measurable functions](/page/Lebesgue%20Integral) and using the Lebesgue integral. The $L^p$ spaces are the result of this construction: they are the completions of spaces of integrable functions under the $p$-th power norm. They form the backbone of modern analysis --- the natural domain and codomain for differential operators in [Sobolev space](/page/Sobolev%20Space) theory, the setting for the [Fourier transform](/page/Fourier%20Transform), and the foundation of probability theory through $L^2(\Omega, \mathcal{F}, \mathbb{P})$.

## Definition

The construction of $L^p$ requires two steps: first defining the *seminormed* space $\mathcal{L}^p$ of $p$-integrable functions, then passing to the quotient by the kernel of the seminorm (functions that are zero almost everywhere). This quotient is essential --- without it, the $L^p$ "norm" would fail to be a true norm, since $\|f\|_p = 0$ would not imply $f = 0$ but only that $f = 0$ $\mu$-almost everywhere.

[definition: $L^p$ Space] Let $(E, \mathcal{E}, \mu)$ be a measure space and let $p \in [1, \infty)$. The **pre-$L^p$ space** $\mathcal{L}^p(E, \mathcal{E}, \mu)$ consists of all $\mathcal{E}$-measurable functions $f: E \to \mathbb{R}$ (or $\mathbb{C}$) satisfying \begin{align*} \|f\|_p := \left( \int_E |f|^p \, d\mu \right)^{1/p} < \infty. \end{align*} For $p = \infty$, the space $\mathcal{L}^\infty(E, \mathcal{E}, \mu)$ consists of all $\mathcal{E}$-measurable functions $f: E \to \mathbb{R}$ satisfying \begin{align*} \|f\|_\infty := \operatorname{ess\,sup}_E |f| := \inf \bigl\{ M \ge 0 : \mu(\{x \in E : |f(x)| > M\}) = 0 \bigr\} < \infty. \end{align*} The functional $\|\cdot\|_p$ is a seminorm on $\mathcal{L}^p$: it satisfies $\|f\|_p \ge 0$, $\|\lambda f\|_p = |\lambda| \|f\|_p$, and the triangle inequality (Minkowski's inequality, stated below), but $\|f\|_p = 0$ forces only $f = 0$ $\mu$-almost everywhere. The **$L^p$ space** is the quotient \begin{align*} L^p(E, \mathcal{E}, \mu) := \mathcal{L}^p(E, \mathcal{E}, \mu) \,/\, \sim, \qquad \text{where } f \sim g \iff f = g \;\; \mu\text{-a.e.} \end{align*} An element of $L^p$ is an [equivalence class](/page/Equivalence%20Relation) $[f]$ of $\mu$-a.e. equal functions. The $L^p$ norm $\|[f]\|_{L^p} := \|f\|_p$ is well-defined on the quotient (independent of representative) and is a true norm. [/definition]

The passage from $\mathcal{L}^p$ to $L^p$ is more than a formality. Working with [equivalence classes](/page/Equivalence%20Relation) means that statements about $L^p$ functions hold only $\mu$-almost everywhere, not pointwise. In particular, "let $f \in L^p(E)$ and suppose $f(x_0) = 3$" is meaningless unless we have a specific continuous representative or a [trace operator](/page/Sobolev%20Space).

[remark: Notation Convention] Throughout this page and on Androma generally, we write $f \in L^p(E)$ rather than $[f] \in L^p(E)$, identifying a function with its equivalence class. When the measure space is $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$ and the context is clear, we abbreviate $L^p(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$ as $L^p(\mathbb{R}^n)$. [/remark]

The following example illustrates how the integrability threshold for $L^p$ depends on the interplay between the singularity strength and the exponent $p$, and reveals the nesting of $L^p$ spaces on bounded domains.

[example: Power Singularities in $L^p(\mathbb{R}^n)$] A basic but instructive class of $L^p$ functions on $\mathbb{R}^n$ consists of radial power functions. Let $U = B(0,1) \subset \mathbb{R}^n$ with $n \ge 1$, and consider the function $f: U \to \mathbb{R}$ defined by $f(x) = |x|^{-\alpha}$ for a parameter $\alpha > 0$. We determine for which values of $p$ and $\alpha$ the function $f$ belongs to $L^p(U)$. Using polar coordinates with $r = |x|$ and the surface area element $d\mathcal{H}^{n-1}$ on the unit sphere $\mathbb{S}^{n-1}$ (where $\mathcal{H}^{n-1}(\mathbb{S}^{n-1}) = n\omega_n$ with $\omega_n$ the volume of the unit ball): \begin{align*} \int_{B(0,1)} |x|^{-\alpha p} \, d\mathcal{L}^n(x) &= n\omega_n \int_0^1 r^{-\alpha p} \cdot r^{n-1} \, dr = n\omega_n \int_0^1 r^{n - 1 - \alpha p} \, dr. \end{align*} The integrand is $r^{n - 1 - \alpha p}$, which is integrable near $r = 0$ if and only if the exponent exceeds $-1$: \begin{align*} n - 1 - \alpha p > -1 \iff \alpha p < n \iff p < \frac{n}{\alpha}. \end{align*} Therefore $|x|^{-\alpha} \in L^p(B(0,1))$ if and only if $p < n/\alpha$. For instance, in dimension $n = 3$: the Coulomb potential singularity $|x|^{-1}$ lies in $L^p(B(0,1))$ for $p < 3$; the stronger singularity $|x|^{-2}$ lies in $L^p(B(0,1))$ only for $p < 3/2$. As $\alpha \to n$, the critical exponent $n/\alpha$ approaches $1$, and the function barely fails to be integrable. This example also reveals the **nesting failure** on bounded domains: higher integrability is a *stronger* condition. If $f \in L^q(U)$ for some $q > p$ and $\mu(E) < \infty$, then $f \in L^p(U)$ (by Holder's inequality, as we show below), but not conversely. [/example]

## The Fundamental Inequalities

Three inequalities form the analytic backbone of $L^p$ theory. Each serves a distinct structural role: Holder's inequality establishes duality, Minkowski's inequality proves the triangle inequality for the $L^p$ norm (making $L^p$ a normed space), and Young's inequality for products is the pointwise engine that drives both.

### Young's Inequality for Products

The proof of Holder's inequality reduces, after normalizing both functions to have unit $L^p$ and $L^q$ norms, to bounding a product $ab$ by a sum of powers $a^p/p + b^q/q$. This reduction is the heart of the argument: if we can control a product by a convex combination of pure powers, then integrating both sides yields the desired integral bound. The need for such a "product-to-sum" conversion arises repeatedly --- in proving Minkowski's inequality, in establishing convolution bounds, and in interpolation estimates --- making the following numerical inequality the load-bearing ingredient of the entire $L^p$ framework.

[quotetheorem:244]

The conjugate exponent relation $1/p + 1/q = 1$ appears throughout $L^p$ theory. Given $p \in (1, \infty)$, we write $q = p/(p-1)$ for its conjugate. The endpoint cases are $p = 1, q = \infty$ and $p = \infty, q = 1$, where the duality pairing takes a different form (see the section on Duality below). The map $p \mapsto q = p/(p-1)$ is a decreasing involution on $(1, \infty)$ that fixes $p = 2$: the space $L^2$ is self-dual, a fact with far-reaching consequences.

### Holder's Inequality