Almost-Everywhere Equivalence in Lp

Why \(L^p\) Uses Equivalence Classes

Lebesgue integration does not detect what a measurable function does on a set of measure zero. Consequently, two functions that differ only on such a set have the same integral of their absolute \(p\)-th powers, and the same \(L^p\) size. The space \(L^p\) is therefore constructed from equivalence classes of functions, not from individual pointwise formulas. This identification is specific to the measure: a set that is null for Lebesgue measure need not be null for counting measure.

Fix a measure space \((X,\mathcal{M},\mu)\), with no assumption that \(\mu\) is finite, sigma-finite, or complete. Functions below are measurable, scalar-valued real or complex functions. For \(1\le p<\infty\), membership in \(L^p(\mu)\) means \(\int_X |f|^p\,d\mu<\infty\), and the usual size is \(\left(\int_X|f|^p\,d\mu\right)^{1/p}\). For \(p=\infty\), membership means that \(f\) is essentially bounded; its usual size is \(\inf\{C\ge0: |f|\le C\ \mu\text{-a.e.}\}\). Here, “a.e.” means outside a measurable set of \(\mu\)-measure zero.

The null set in this definition is measurable: for measurable real- or complex-valued functions, the set where \(f\ne g\) is measurable. If the measure space is not complete, an arbitrary modification on a subset of a null set need not be measurable. Thus the definition compares measurable functions; it does not assert that every pointwise alteration is measurable.

Null Sets Do Not Change Integrals or Sizes

Proof. Let \(N=\{x:f(x)\ne g(x)\}\), so \(\mu(N)=0\). For finite \(p\), the nonnegative measurable functions \(|f|^p\) and \(|g|^p\) agree on \(N^c\). The integral of any nonnegative measurable function supported on \(N\) is zero: every nonnegative simple function below it is supported on \(N\), so each of its positive-level sets has measure zero; the definition of the nonnegative integral as a supremum over such simple functions gives integral zero. Splitting each power into its parts on \(N\) and \(N^c\), additivity of the nonnegative integral gives \[ \int_X|f|^p\,d\mu=\int_{N^c}|f|^p\,d\mu =\int_{N^c}|g|^p\,d\mu=\int_X|g|^p\,d\mu. \] Thus one integral is finite exactly when the other is, and their \(p\)-th roots agree.

For \(p=\infty\), fix \(C\ge0\). The conditions \(|f|\le C\) a.e. and \(|g|\le C\) a.e. are equivalent: outside the union of \(N\) and the exceptional null set for either bound, \(f=g\), so either bound transfers to the other function. Hence the sets of admissible constants \(C\) in the two essential-supremum definitions are identical. Their infima agree, and one function is essentially bounded exactly when the other is. \(\square\)

This theorem makes the usual \(L^p\) size a function of the class \([f]\), rather than of the representative selected from that class. In particular, writing \(\|[f]\|_p\) for the common value is unambiguous.

Building the Quotient Space

To construct \(L^p\), we first verify that almost-everywhere equality really is an equivalence relation. We also need addition and scalar multiplication to make sense on classes: the result must not depend on which representatives are chosen. The next proposition establishes these facts for the usual range \(1\le p\le\infty\).

Proof. Reflexivity holds because \(\{f\ne f\}=\varnothing\), and \(\mu(\varnothing)=0\). Symmetry holds because \(\{f\ne g\}=\{g\ne f\}\). If \(f\sim g\) and \(g\sim h\), then \(\{f\ne h\}\subseteq\{f\ne g\}\cup\{g\ne h\}\), a union of two null sets. Therefore \(f\sim h\), proving transitivity.

Suppose \(f\sim \tilde f\) and \(g\sim\tilde g\). Outside the union of the two null sets where these equalities can fail, \(f+g=\tilde f+\tilde g\) and \(af=a\tilde f\). Thus the formulas for addition and scalar multiplication do not depend on the representatives.

It remains to check membership. For \(1\le p<\infty\), the scalar inequality \[ |z+w|^p\le 2^{p-1}(|z|^p+|w|^p) \] holds for real or complex \(z,w\): the triangle inequality bounds \(|z+w|\) by \(|z|+|w|\), and convexity of \(t\mapsto t^p\) gives \(((u+v)/2)^p\le (u^p+v^p)/2\) for \(u,v\ge0\). Applying this inequality pointwise and integrating shows \(\int|f+g|^p\,d\mu\le2^{p-1}(\int|f|^p\,d\mu+\int|g|^p\,d\mu)<\infty\). Also \(\int|af|^p\,d\mu=|a|^p\int|f|^p\,d\mu<\infty\). For \(p=\infty\), if \(|f|\le C\) and \(|g|\le D\) a.e., then \(|f+g|\le C+D\) a.e. and \(|af|\le |a|C\) a.e.; hence the sum and scalar multiple are essentially bounded. The zero class and additive inverses are represented by the zero function and \(-f\), respectively. This proves the asserted vector-space operations. \(\square\)

The quotient construction identifies functions that differ only on null sets; it does not claim that equivalent functions agree at every point. Nor does this proposition establish the triangle inequality for the induced \(L^p\) size: that norm property is treated separately in the study of the basic \(L^p\) inequalities.

Zero Size and Equality Almost Everywhere

The quotient also explains why a zero \(L^p\) size means equality almost everywhere, not necessarily equality everywhere. For \(1\le p<\infty\), if \(\int|f|^p\,d\mu=0\), then for each positive integer \(n\), the nonnegative simple function \(n^{-p}\chi_{\{|f|>1/n\}}\) is bounded above by \(|f|^p\). Its integral is therefore zero, so \(\mu(\{|f|>1/n\})=0\). Since \(\{|f|>0\}=\bigcup_{n\ge1}\{|f|>1/n\}\), countable subadditivity gives \(f=0\) a.e. Conversely, if \(f=0\) a.e., the integral of \(|f|^p\) is zero by the null-set argument in the theorem.

For \(p=\infty\), if \(\|f\|_\infty=0\), then for every \(n\ge1\) there is an admissible essential bound \(C_n<1/n\), by the definition as an infimum. Outside a null set, \(|f|\le C_n<1/n\); taking the union of these countably many exceptional null sets shows \(|f|=0\) a.e. Conversely, a function equal to zero a.e. has essential bound \(0\). Applying these observations to \(f-g\) shows that zero \(L^p\) distance corresponds exactly to \(f=g\) a.e.

Common pitfall. A pointwise change is not automatically an \(L^p\) change. For the first example, the incorrect calculation \(\int_0^1g^2\,dx=\int_0^1x^2\,dx+100^2\) treats the point \(1/2\) as if it had positive measure. The correction is that the altered value contributes only on a singleton of measure zero, so \(\int_0^1g^2\,dx=1/3\). Conversely, under counting measure a single point has measure \(1\), not zero, so changing one value can change both the class and the integral.