The first reason to invent $L^p$ spaces is that pointwise control is too rigid for many analytic problems: an integral can stay small while a function develops narrow spikes, and a sequence of functions can converge in average size while refusing to converge at any fixed point. The $L^p$ viewpoint measures the size of a function by integrating a power of its magnitude, then treats functions that differ only on a null set as the same analytic object.

This change of viewpoint is not cosmetic. It is the move that lets limits, approximation, Fourier analysis, probability, and partial differential equations live in common function spaces. Instead of asking whether $f_n(x)$ approaches $f(x)$ for every $x$, we ask whether the total $p$-power error tends to zero. That question is stable under the operations that analysis uses most often.

[example: Narrow Spikes With Small $L^p$ Mass] Let $X=[0,1]$ with [Lebesgue measure](/page/Lebesgue%20Measure), fix $1\leq p<\infty$, and define \begin{align*} f_n(x)=n^{1/p}\mathbb{1}_{(0,1/n)}(x). \end{align*} For $x\in(0,1/n)$, one has $|f_n(x)|^p=(n^{1/p})^p=n$, while for $x\notin(0,1/n)$ one has $|f_n(x)|^p=0$. Hence \begin{align*} \int_0^1 |f_n(x)|^p\,dx=\int_0^{1/n} n\,dx. \end{align*} Since the integrand is constant on $(0,1/n)$, \begin{align*} \int_0^{1/n} n\,dx=n\left(\frac{1}{n}-0\right)=1. \end{align*} Therefore $\|f_n\|_p^p=1$, and so $\|f_n\|_p=1$ for every $n$. Now define \begin{align*} g_n(x)=n^{1/p}\mathbb{1}_{(0,1/n^2)}(x). \end{align*} For $x\in(0,1/n^2)$, one has $|g_n(x)|^p=n$, and outside this interval the value is $0$. Thus \begin{align*} \|g_n\|_p^p=\int_0^1 |g_n(x)|^p\,dx=\int_0^{1/n^2} n\,dx. \end{align*} Again using constancy of the integrand, \begin{align*} \int_0^{1/n^2} n\,dx=n\left(\frac{1}{n^2}-0\right)=\frac{1}{n}. \end{align*} Thus $\|g_n\|_p^p=1/n$ and $\|g_n\|_p=n^{-1/p}\to 0$. The two families both form tall spikes near $0$, but $L^p$ size records the product of height-to-the-$p$ and supporting measure, not the visual height alone. [/example]

The example shows the central tradeoff. $L^p$ norms do not remember where every value occurs, but they remember how much $p$-power is distributed across the underlying [measure space](/page/Measure%20Space). That is exactly the information needed for integral estimates.

## Definition

To make the preceding measurement into a space, we must decide two things: which functions have finite size, and when two functions should be identified. The null-set identification is forced by integration, since the integral cannot distinguish functions that disagree only on a set of measure zero.

[definition: $L^p$ Space] Let $(X,\mathcal{A},\mu)$ be a measure space, let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$, and let $1 \leq p \leq \infty$. For $1 \leq p < \infty$, $L^p(X,\mathcal{A},\mu;\mathbb{F})$ is the set of equivalence classes of [measurable functions](/page/Measurable%20Functions) $f: X \to \mathbb{F}$ such that \begin{align*} \int_X |f|^p \, d\mu < \infty. \end{align*} For $p=\infty$, $L^\infty(X,\mathcal{A},\mu;\mathbb{F})$ is the set of equivalence classes of measurable functions $f: X \to \mathbb{F}$ for which there exists $M \in [0,\infty)$ such that $|f(x)| \leq M$ for $\mu$-almost every $x \in X$. Two measurable functions represent the same element when they agree $\mu$-almost everywhere. For $1 \leq p < \infty$, the norm is the map $\|\cdot\|_p: L^p(X,\mathcal{A},\mu;\mathbb{F}) \to [0,\infty)$ given by \begin{align*} \|f\|_p = \left(\int_X |f|^p \, d\mu\right)^{1/p}. \end{align*} For $p=\infty$, the norm is the map $\|\cdot\|_\infty: L^\infty(X,\mathcal{A},\mu;\mathbb{F}) \to [0,\infty)$ given by \begin{align*} \|f\|_\infty = \inf\{M \in [0,\infty): |f(x)| \leq M \text{ for } \mu\text{-almost every } x \in X\}. \end{align*} [/definition]

After passing to equivalence classes, the notation $f(x)$ is a convenient representative-level phrase rather than a value attached to the equivalence class. This convention is harmless as long as statements are invariant under changes on null sets. It is also the source of many beginner mistakes: an element of $L^p$ need not have a preferred value at a point.

[remark: Naming the Underlying Measure] When the measure space is understood, analysts write $L^p(X)$ or $L^p(\mu)$. On a domain $\Omega \subset \mathbb{R}^n$, the default usually means $L^p(\Omega,\mathcal{B}(\Omega),\mathcal{L}^n;\mathbb{F})$, where $\mathcal{L}^n$ is Lebesgue measure. [/remark]

The definition includes several familiar spaces at once. For counting measure on $\mathbb{N}$ it produces sequence spaces; for [probability measure](/page/Probability%20Measure) it describes random variables with finite moments; for Lebesgue measure on a domain it gives the standard function spaces of analysis.

## Integral Size and First Examples

The same formula behaves differently on finite, infinite, atomic, and non-atomic measure spaces. This section records the main tests that tell the reader what the norm is actually seeing.

### Sequence Spaces

Counting measure turns integration into summation, so $L^p$ becomes the space where the $p$-power sum converges. This example is the cleanest way to see that $p$ controls a tail condition, not merely a pointwise bound.

[example: The Sequence Space $\ell^p$] Let $X=\mathbb{N}$, let $\mathcal{A}=2^{\mathbb{N}}$, and let $\mu$ be counting measure. A [measurable function](/page/Measurable%20Function) $a:\mathbb{N}\to\mathbb{F}$ is the same thing as a sequence $a=(a_n)_{n=1}^{\infty}$, because every subset of $\mathbb{N}$ is measurable. For $1\leq p<\infty$, integration with respect to counting measure gives \begin{align*} \int_{\mathbb{N}} |a|^p\,d\mu=\sum_{n=1}^{\infty} |a(n)|^p=\sum_{n=1}^{\infty}|a_n|^p. \end{align*} Therefore $a\in L^p(\mathbb{N},2^{\mathbb{N}},\mu;\mathbb{F})$ exactly when \begin{align*} \sum_{n=1}^{\infty}|a_n|^p<\infty. \end{align*} In that case, \begin{align*} \|a\|_p=\left(\sum_{n=1}^{\infty}|a_n|^p\right)^{1/p}. \end{align*} For $p=\infty$, the only counting-measure null set is $\varnothing$, since every nonempty subset of $\mathbb{N}$ has positive counting measure. Thus $|a_n|\leq M$ for almost every $n$ is the same as $|a_n|\leq M$ for every $n$, and hence \begin{align*} \|a\|_\infty=\inf\{M\in[0,\infty): |a_n|\leq M\text{ for every }n\in\mathbb{N}\}=\sup_{n\in\mathbb{N}}|a_n|. \end{align*} So membership in $L^\infty$ is exactly boundedness of the sequence. For the concrete sequence $a_n=n^{-\alpha}$ with $\alpha>0$ and $1\leq p<\infty$, we have \begin{align*} |a_n|^p=|n^{-\alpha}|^p=n^{-\alpha p}. \end{align*} Thus \begin{align*} \sum_{n=1}^{\infty}|a_n|^p=\sum_{n=1}^{\infty}n^{-\alpha p}. \end{align*} By the *[p-series test](/theorems/3834)*, the series $\sum_{n=1}^{\infty}n^{-s}$ converges exactly when $s>1$. Applying this with $s=\alpha p$, the sequence $(n^{-\alpha})$ belongs to $\ell^p$ exactly when $\alpha p>1$. For example, if $\alpha=3/4$, then $\alpha\cdot 2=3/2>1$ but $\alpha\cdot 1=3/4\leq 1$, so $(n^{-3/4})\in\ell^2$ and $(n^{-3/4})\notin\ell^1$. The exponent $p$ therefore controls how much tail decay is required, not merely whether the terms are individually small. [/example]

Sequence spaces show why no single exponent is universally best. Larger $p$ penalizes large terms more strongly, while smaller $p$ is more sensitive to long tails. On an infinite measure space these pressures are not ordered in a simple inclusion chain.

### Finite Measure Spaces

On a finite measure space the total amount of room is bounded, so high integrability forces low integrability. This is a first example of a recurring theme: the measure of the ambient space matters as much as the formula.

[quotetheorem:9904]

The theorem explains why probability theory often treats higher moments as stronger information. When $\mu(X)=1$, the estimate becomes $\|f\|_p \leq \|f\|_q$. The statement fails on many infinite measure spaces, so finite measure is not a decorative hypothesis.