Why Lp Inclusions Fail on Infinite-Measure Spaces

Why Infinite Measure Changes the Picture

On a finite-measure space, Hölder's inequality gives an inclusion from \(L^q\) into \(L^p\) when \(1\le p<q<\infty\). On \(\mathbb{R}\) with Lebesgue measure, the corresponding global inclusion fails: there are functions in \(L^p\) but not \(L^q\), and other functions in \(L^q\) but not \(L^p\). The key is that a function can take different sizes on disjoint sets whose measures also vary. The sizes control the powers of the function; the set measures let us tune the contributions to the two integrals independently.

Throughout the main construction, \(1\le p<q<\infty\), and the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. Functions are real-valued and measurable. For \(1\le r<\infty\), \(L^r(m)\) consists of equivalence classes, modulo equality \(m\)-a.e., of functions \(f\) for which \(\int_{\mathbb{R}}|f|^r\,dm<\infty\); the norm is \(\|f\|_r=(\int_{\mathbb{R}}|f|^r\,dm)^{1/r}\).

The finite-measure comparison explains why the issue arises. On a measurable set \(A\) with \(m(A)<\infty\), Hölder's inequality gives \(\|f\|_{L^p(A)}\le m(A)^{1/p-1/q}\|f\|_{L^q(A)}\). That factor depends on \(m(A)\). Taking \(A=\mathbb{R}\) is not valid, because \(m(\mathbb{R})=\infty\). The following construction shows that this is a genuine obstruction, not merely a limitation of that estimate.

A Disjoint-Support Calculation

Proof. For each \(N\), set \(f_N=\sum_{n=1}^{N}c_n\chi_{E_n}\). Since the sets are disjoint, at most one summand is nonzero at any point, so \(f_N^r=\sum_{n=1}^{N}c_n^r\chi_{E_n}\). Integrating this finite sum gives \(\int f_N^r\,dm=\sum_{n=1}^{N}c_n^r m(E_n)\). The functions \(f_N^r\) are nonnegative, measurable, and increase pointwise to \(f^r\). The Monotone Convergence Theorem therefore gives \(\int f^r\,dm=\lim_{N\to\infty}\int f_N^r\,dm=\sum_{n=1}^{\infty}c_n^r m(E_n)\). This also covers zero coefficients and sets of measure zero. \(\square\)

Thus membership in \(L^r(m)\) for one of these functions reduces to convergence of the corresponding numerical series. We use disjoint half-open intervals to realize the required measures. Given any positive finite lengths \(\ell_n\), the intervals \([s_n,s_n+\ell_n)\), where \(s_1=0\) and \(s_n=\sum_{k=1}^{n-1}\ell_k\) for \(n\ge2\), are pairwise disjoint, measurable, and have measures \(\ell_n\). Each starting point is finite because it is a finite sum of finite lengths.

Two Constructions for Any Finite Exponents

Proof. Write \(d=q-p>0\). For the first function, choose pairwise disjoint half-open intervals \(E_n\) with \(m(E_n)=n^{-2-p/d}\), and set \(f=\sum_{n=1}^{\infty}n^{1/d}\chi_{E_n}\). These lengths and values are finite for each \(n\), and the function is measurable as a pointwise limit of measurable partial sums. The lemma gives, term by term,

Here \(q/d-p/d=1\). The series with terms \(1/n^2\) converges, for example by comparison with \(\int_1^\infty x^{-2}\,dx\). The harmonic series diverges: in the block \(2^k\le n<2^{k+1}\), there are \(2^k\) terms, each at least \(1/2^{k+1}\), so each block contributes at least \(1/2\). Consequently \(f\in L^p(m)\) and \(f\notin L^q(m)\).

For the reverse direction, choose pairwise disjoint half-open intervals \(F_n\) with \(m(F_n)=n^{-2+q/d}\), and set \(g=\sum_{n=1}^{\infty}n^{-1/d}\chi_{F_n}\). Each interval still has a positive finite length, even if the lengths grow with \(n\). The lemma now yields

In the last exponent, \((q-p)/d=1\). The same convergence and divergence checks apply. Thus \(g\in L^q(m)\) and \(g\notin L^p(m)\). Both functions are finite-valued at every point: disjointness means each point belongs to at most one interval. This proves both failures. \(\square\)

What the Examples Do—and Do Not—Say

The examples show that on \((\mathbb{R},\mathcal{L},m)\) neither inclusion \(L^p(m)\subseteq L^q(m)\) nor \(L^q(m)\subseteq L^p(m)\) holds when \(1\le p<q<\infty\). They do not say that every function changes membership when the exponent changes. For instance, on \(A=[0,1)\), the function \(\chi_A\) belongs to both spaces: for either finite \(r\ge1\), \(\int_{\mathbb{R}}|\chi_A|^r\,dm=m(A)=1\). The construction needs a deliberate balance between set size and function height.

Common mistake: use unit intervals for both directions. A tempting attempt for \(L^2\) but not \(L^4\) is to put the value \(n^{-3/8}\) on each unit interval \([n-1,n)\). The \(L^2\) integral then equals \(\sum n^{-3/4}\), which diverges, while the \(L^4\) integral equals \(\sum n^{-3/2}\), which converges. This gives the reverse example, not the intended one. The correction is to vary the interval measures as well as the values. In the first worked example, lengths \(n^{-3}\) and values \(n^{1/2}\) give \(L^2\) contributions \(n^{-2}\) and \(L^4\) contributions \(n^{-1}\), producing the desired direction. Unit intervals are useful for the reverse direction, but alone they cannot produce both failures.

If one exponent is \(\infty\), the same absence of global inclusion has simple endpoint examples. For any finite \(1\le p<\infty\), the constant function \(1\) on \(\mathbb{R}\) is essentially bounded, so it lies in \(L^\infty(m)\), but \(\int_{\mathbb{R}}1^p\,dm=m(\mathbb{R})=\infty\), so it is not in \(L^p(m)\). Conversely, define \(h(x)=x^{-1/(2p)}\) for \(0<x<1\) and \(h(x)=0\) elsewhere. Then \(\int|h|^p\,dm=\int_0^1x^{-1/2}\,dx=2<\infty\), while \(h\) is not essentially bounded: for every \(K>0\), the interval \(0<x<K^{-2p}\) (when \(K\ge1\)) has positive measure and \(h(x)>K\) there. Thus \(h\in L^p(m)\) but \(h\notin L^\infty(m)\).