From Integrability to \(L^p\) Spaces
Earlier in the course, integrability meant that the integral of the absolute value of a measurable function is finite. The spaces \(L^p\) extend this idea: instead of measuring the total size of \(|f|\), we measure the total size of a positive power \(|f|^p\). The choice of \(p\) changes which functions are admitted. A function with a strong but localized singularity, for example, may belong to some \(L^p\) spaces and not to others.
There is an important qualification. Lebesgue integration does not distinguish functions that differ only on a set of measure zero. Accordingly, the elements of \(L^p\) are not individual functions but equivalence classes of functions that agree almost everywhere. We first make this precise, then establish a level-set estimate and compare different exponents on finite-measure spaces. The next tutorial develops the norm associated with these spaces.
Definition of \(L^p\)
Fix a measure space \((X,\mathcal{F},\mu)\). For a finite exponent \(p\geq1\), a measurable real-valued function belongs to \(L^p\) when its \(p\)-th absolute moment has finite integral. This definition uses the integral of a nonnegative measurable function, so it also applies when the measure of \(X\) is infinite.
The expression “equal almost everywhere” means that the set \(\{x:f(x)\neq g(x)\}\) has measure zero. The earlier theorem on invariance of the integral under almost-everywhere equality ensures that changing a representative on a null set does not change the integral that defines membership. The same idea applies at the endpoint \(p=\infty\), where membership is described by an almost-everywhere bound rather than by an integral of a finite power.
Essential boundedness is not the same as boundedness at every point. A function may take an unusually large value on a null set and still be essentially bounded. As with finite \(p\), this definition is deliberately made on equivalence classes: changing values on a null set does not change membership in \(L^\infty\).
Worked Examples of Membership
Worked Example: A Power Singularity on a Finite Interval
Let \(X=(0,1)\) with Lebesgue measure, and let \(f(x)=x^{-\alpha}\), where \(\alpha>0\). For any \(p\geq1\), $$ \int_0^1 |f(x)|^p\,dx =\int_0^1 x^{-\alpha p}\,dx. $$ If \(\alpha p\neq1\), integrating from \(a\) to \(1\) and then taking \(a\downarrow0\) gives $$ \int_a^1 x^{-\alpha p}\,dx =\frac{1-a^{1-\alpha p}}{1-\alpha p}. $$ When \(\alpha p<1\), \(a^{1-\alpha p}\to0\), so the integral is \(1/(1-\alpha p)\), which is finite. When \(\alpha p>1\), \(a^{1-\alpha p}\to\infty\), so the integral diverges. In the remaining case \(\alpha p=1\), $$ \int_a^1 x^{-1}\,dx=-\log a\longrightarrow\infty. $$ Thus \(f\in L^p(0,1)\) exactly when \(\alpha p<1\). For instance, with \(\alpha=1/3\), the function belongs to \(L^2(0,1)\), since \(2/3<1\), but not to \(L^3(0,1)\), since \(3/3=1\).
Worked Example: A Function in Every Finite \(L^p\) but Not in \(L^\infty\)
On \((0,1)\), again with Lebesgue measure, set \(g(x)=|\log x|\). For each finite \(p\geq1\), use the substitution \(t=-\log x\), so \(x=e^{-t}\) and \(dx=-e^{-t}\,dt\). Reversing the limits gives $$ \int_0^1|\log x|^p\,dx =\int_0^\infty t^p e^{-t}\,dt. $$ This integral is finite: near \(t=0\), \(t^pe^{-t}\leq t^p\), which is integrable on \((0,1)\); for sufficiently large \(t\), \(p\log t\leq t/2\), and therefore \(t^pe^{-t}\leq e^{-t/2}\), which is integrable on a tail. Hence \(g\in L^p(0,1)\) for every finite \(p\geq1\).
However, \(g\) is not essentially bounded. Given any finite \(M\geq0\), the set where \(|\log x|>M\) contains \((0,e^{-M})\), whose measure is \(e^{-M}>0\). Thus no finite bound holds outside a null set, and \(g\notin L^\infty(0,1)\). This shows why membership in every finite-\(p\) space need not imply essential boundedness.
Worked Example: A Null-Set Modification and Essential Boundedness
Define \(h:\mathbb{R}\to\mathbb{R}\) by \(h(0)=10^6\) and \(h(x)=0\) for \(x\neq0\), with Lebesgue measure on \(\mathbb{R}\). The function is measurable and differs from the zero function only on the singleton \(\{0\}\), which has measure zero. For every finite \(p\geq1\), $$ \int_{\mathbb{R}}|h|^p\,dx =(10^6)^p\,\mu(\{0\})=0. $$ It follows that \(h\) represents the zero element of \(L^p(\mathbb{R})\). It is also essentially bounded: for example, with \(M=0\), the set \(\{|h|>0\}\) is \(\{0\}\) and has measure zero. Although \(h\) is not bounded by zero at every point, it belongs to \(L^\infty(\mathbb{R})\).
A Level-Set Estimate
Membership in \(L^p\) controls not only the integral of \(|f|^p\), but also the measure of the set where \(|f|\) exceeds a chosen threshold. The estimate below follows directly by comparing the integrand with a constant on that set. It is often called the Markov inequality for functions.
Proof. On the set \(A=\{x:|f(x)|>t\}\), the inequality \(|f(x)|^p\geq t^p\) holds. Therefore, pointwise on \(X\), $$ t^p\mathbf{1}_A\leq |f|^p. $$ Both sides are nonnegative and measurable. By monotonicity of the Lebesgue integral and the integral-of-an-indicator theorem, $$ t^p\mu(A) =\int_X t^p\mathbf{1}_A\,d\mu \leq\int_X|f|^p\,d\mu. $$ The right-hand side is finite by the definition of \(L^p\), and \(t^p>0\). Dividing by \(t^p\) proves the estimate. In particular, every such level set has finite measure, even when \(\mu(X)\) is infinite.
The estimate gives a useful way to read the \(L^p\) condition: larger thresholds force the measure of the corresponding level sets to be small. It does not say that \(f\) is bounded. In the power-singularity example, \(f\) is unbounded, but it still belongs to some finite-\(p\) spaces because the sets where it is large are sufficiently small in the integral sense.
Basic Closure and Comparison of Spaces
The terminology “space” is justified in part because these classes are closed under the usual linear operations. We establish this without using the \(L^p\) norm. For \(p\geq1\), convexity of \(u\mapsto u^p\) on \([0,\infty)\) gives, for nonnegative \(u,v\), $$ \left(\frac{u+v}{2}\right)^p\leq\frac{u^p+v^p}{2}. $$ Combining this with \(|f+g|\leq|f|+|g|\) yields the needed integrable upper bound.
Proof. For finite \(p\), the convexity inequality above implies $$ |f+g|^p\leq (|f|+|g|)^p \leq 2^{p-1}\bigl(|f|^p+|g|^p\bigr). $$ The right-hand side has finite integral because \(f\) and \(g\) belong to \(L^p\). Thus \(f+g\in L^p\). For a scalar \(a\), \(|af|^p=|a|^p|f|^p\), so \(af\in L^p\); the case \(a=0\) is included. Applying these facts to \(af\) and \(bg\) proves \(af+bg\in L^p\).
For \(L^\infty\), choose finite constants \(M,N\) such that \(|f|\leq M\) and \(|g|\leq N\) almost everywhere. Outside the union of the two exceptional null sets, \(|af+bg|\leq |a|M+|b|N\). A finite union of null sets is null, so \(af+bg\) is essentially bounded. This proves closure for \(L^\infty\) as well.
On a finite-measure space, a larger finite exponent implies membership at every smaller exponent. This inclusion can be proved by splitting the domain according to whether \(|f|\) is at most one. No inequality between integral norms is needed.
Proof. Partition \(X\) into \(E=\{|f|\leq1\}\) and \(F=\{|f|>1\}\). On \(E\), \(|f|^p\leq1\). On \(F\), because \(q>p\) and \(|f|>1\), \(|f|^p\leq|f|^q\). Consequently, $$ \int_X|f|^p\,d\mu =\int_E|f|^p\,d\mu+\int_F|f|^p\,d\mu \leq\mu(E)+\int_F|f|^q\,d\mu \leq\mu(X)+\int_X|f|^q\,d\mu. $$ Both terms on the final line are finite. Therefore \(f\in L^p(X,\mu)\), as claimed.
Worked Example: Why Finite Measure Matters for Inclusion
On \(\mathbb{R}\) with Lebesgue measure, let \(u(x)=(1+|x|)^{-1/2}\). For \(x\geq0\), \(u(x)^2=(1+x)^{-1}\), whose integral on \([0,\infty)\) diverges. By symmetry, \(\int_{\mathbb{R}}|u|^2\,dx=\infty\), so \(u\notin L^2(\mathbb{R})\).
For \(x\geq0\), \(u(x)^4=(1+x)^{-2}\), and $$ \int_{\mathbb{R}}|u|^4\,dx =2\int_0^\infty(1+x)^{-2}\,dx =2. $$ Thus \(u\in L^4(\mathbb{R})\) but \(u\notin L^2(\mathbb{R})\). This does not contradict the finite-measure inclusion theorem: \(\mathbb{R}\) has infinite measure. The theorem’s finite-measure hypothesis is essential to its general conclusion.
What the Definitions Do—and Do Not—Say
For finite \(p\), membership is an integral condition, not a pointwise bound. A function may become arbitrarily large and still belong to \(L^p\), provided its \(p\)-th power has finite integral. Conversely, a function that is small at each point in some informal sense may fail to belong to \(L^p\) on an infinite-measure space if that small contribution persists across too much of the domain.
The endpoint \(L^\infty\) is different: it asks for a bound outside a null set, rather than finiteness of an integral of a finite power. Also, the finite-measure inclusion result should not be applied without checking \(\mu(X)<\infty\). The last example shows that on an infinite-measure space, \(L^q\) need not be contained in \(L^p\) when \(p<q\).
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why is a function changed on a null set regarded as the same element of \(L^p\)?
- For \(f(x)=x^{-\alpha}\) on \((0,1)\), what condition on \(\alpha\) and \(p\) makes \(f\) belong to \(L^p(0,1)\)?
- How does the level-set estimate bound \(\mu(\{|f|>t\})\) in terms of \(t\) and the integral of \(|f|^p\)?
- Which step in the proof that \(L^q\subseteq L^p\) uses the assumption that the whole space has finite measure?
- Why does \(u(x)=(1+|x|)^{-1/2}\) on \(\mathbb{R}\) show that finite-measure inclusion cannot be applied to arbitrary measure spaces?