From Measurable Functions to \(L^p\)
Measurability makes integration possible, but it does not guarantee that the integral of a function's size is finite. The spaces \(L^p\) sort measurable functions according to whether their \(p\)-th power has finite integral. The measure matters: an integral on the real line, a sum over the natural numbers, and an integral against a measure concentrated at one point test functions in different ways.
Fix a measure space \((X,\mathcal{M},\mu)\), with no assumption that \(\mu\) is finite, sigma-finite, or complete. Let \(p\) be a real number with \(1\le p<\infty\). Functions in the definition below are \(\mathcal{M}\)-measurable and real-valued or complex-valued. The integrand \(|f|^p\) is nonnegative, so its Lebesgue integral is defined, possibly with value \(+\infty\).
The finiteness condition is about the integral, not whether \(f\) is bounded or finite at each point. A function can be unbounded and still be \(p\)-integrable; a bounded function on a space of infinite measure need not be. A null set is a measurable set of measure zero. Changing a function on a null set does not change the integral of \(|f|^p\), which is why the definition identifies functions equal almost everywhere. The details of this identification are a subject in their own right; here, membership calculations can be made using any representative.
A Basic Bound on Finite Measure Spaces
Proof. The pointwise bound gives \(0\le |f|^p\le M^p\). By monotonicity of the integral on nonnegative measurable functions, \(\int_X|f|^p\,d\mu\le\int_X M^p\,d\mu=M^p\mu(X)\), which is finite. This includes \(M=0\), when the integral is zero, and \(\mu(X)=0\), when the upper bound and integral are both zero. The conclusion holds for every finite \(p\ge1\). \(\square\)
Both hypotheses matter. Finiteness of the measure allows the constant function \(M^p\) to have finite integral, while boundedness supplies the pointwise comparison. The proposition does not assert that bounded measurable functions belong to \(L^p\) when \(\mu(X)=\infty\).
Worked Example: A Piecewise Constant Function on a Finite Interval
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra on \([0,1]\), and let \(\mu\) be Lebesgue measure. Define \(f(x)=2\) for \(0\le x<1/3\), and \(f(x)=-1\) for \(1/3\le x\le1\). Determine its \(L^p\) integral for any \(1\le p<\infty\).
The two intervals have measures \(1/3\) and \(2/3\), respectively. Their endpoint convention assigns \(x=1/3\) to the second interval, and the singleton endpoint has measure zero. Since \(|2|^p=2^p\) and \(|-1|^p=1\), integration of this simple function gives $$ \int_{[0,1]}|f|^p\,dm =2^p\cdot\frac13+1\cdot\frac23 =\frac{2^p+2}{3}. $$
This quantity is finite for every finite \(p\ge1\), so \(f\in L^p([0,1],m)\) for all such \(p\). For \(p=1\), the formula gives \((2+2)/3=4/3\); for \(p=2\), it gives \((4+2)/3=2\). Check: directly at \(p=2\), the squared magnitudes are \(4\) on a set of measure \(1/3\) and \(1\) on a set of measure \(2/3\), so the integral is \(4/3+2/3=2\), as calculated.
Lebesgue Measure on the Real Line
Worked Example: An Exponential on the Real Line
Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. For \(f(x)=e^{-|x|}\), compute the \(L^p\) integral for \(1\le p<\infty\).
The function is nonnegative and measurable, and \(|f(x)|^p=e^{-p|x|}\). Splitting the integral at zero and using the evenness of the integrand gives $$ \int_{\mathbb{R}}|f|^p\,dm =\int_{-\infty}^0e^{px}\,dx+\int_0^\infty e^{-px}\,dx. $$
Because \(p\ge1\), in particular \(p>0\), both improper integrals converge. Evaluating them, \(\int_{-\infty}^0e^{px}\,dx=[e^{px}/p]_{-\infty}^0=1/p\) and \(\int_0^\infty e^{-px}\,dx=[-e^{-px}/p]_0^\infty=1/p\). Thus $$ \int_{\mathbb{R}}|f|^p\,dm=\frac{2}{p}<\infty. $$
Therefore \(f\in L^p(\mathbb{R},m)\) for every finite \(p\ge1\). Check: at \(p=1\), the two half-line integrals are each \(1\), giving \(2\); at \(p=2\), each is \(1/2\), giving \(1\), in agreement with \(2/p\).
This example also shows why an infinite measure space does not rule out \(L^p\) membership: the function decreases sufficiently for its \(p\)-th power to have finite integral. In contrast, the constant function \(1\) on \(\mathbb{R}\) has \(\int_{\mathbb{R}}|1|^p\,dm=m(\mathbb{R})=\infty\), so it is not in \(L^p(\mathbb{R},m)\) for any finite \(p\).
Counting Measure Turns the Integral into a Series
On a countable set \(X\), counting measure assigns to each set its number of elements, with infinite value for an infinite set. Its measurable sets are all subsets of \(X\). For a nonnegative function \(g\), its integral is the sum of its values, with \(+\infty\) allowed. Consequently, membership in \(L^p\) is a summability condition on the sequence of \(p\)-th powers.
Worked Example: A Geometric Sequence on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure. Set \(a_n=3^{-n}\). For \(1\le p<\infty\), calculate the integral of the \(p\)-th power of the function \(a(n)=a_n\).
By the definition of integration with counting measure, $$ \int_{\mathbb{N}}|a|^p\,d\mu =\sum_{n=1}^\infty |3^{-n}|^p =\sum_{n=1}^\infty (3^{-p})^n. $$
Here \(0<3^{-p}<1\). The finite geometric sum through \(n=N\) is \(\sum_{n=1}^N(3^{-p})^n=3^{-p}(1-(3^{-p})^N)/(1-3^{-p})\). As \(N\to\infty\), \((3^{-p})^N\to0\), so $$ \int_{\mathbb{N}}|a|^p\,d\mu =\frac{3^{-p}}{1-3^{-p}} =\frac{1}{3^p-1}<\infty. $$
Thus \(a\in L^p(\mathbb{N},\mu)\) for every finite \(p\ge1\). Check: when \(p=1\), the integral is \(1/(3-1)=1/2\), and the series is \(1/3+1/9+1/27+\cdots=1/2\). For contrast, the constant sequence \(b_n=1\) has \(\int_{\mathbb{N}}|b|^p\,d\mu=\sum_{n=1}^\infty1=\infty\), so it is not in \(L^p(\mathbb{N},\mu)\).
A Dirac Measure Tests One Point
For \(x_0\in X\), the Dirac measure \(\delta_{x_0}\) is defined on a measurable space \((X,\mathcal{M})\) by \(\delta_{x_0}(E)=1\) if \(x_0\in E\), and \(0\) if \(x_0\notin E\). For measurable real- or complex-valued \(f\), integration against this measure gives the value at \(x_0\). This follows first for indicator functions from the definition, then for nonnegative simple functions by finite addition, and for nonnegative measurable functions by the definition of the Lebesgue integral as a supremum of simple-function integrals. Applying it to \(|f|^p\) gives the following result.
Proof. The integral of a nonnegative measurable function \(g\) against \(\delta_{x_0}\) is \(g(x_0)\): for an indicator this is exactly \(\delta_{x_0}(E)=\chi_E(x_0)\); for a nonnegative simple function it follows by summing its indicator terms; and for general \(g\) it follows by taking the supremum over simple functions bounded above by \(g\). Take \(g=|f|^p\). Since \(f(x_0)\) is finite, \(|f(x_0)|^p<\infty\), which proves both the formula and membership. \(\square\)
Worked Example: Functions Under a Dirac Measure
Take \(X=\mathbb{R}\) with its Borel sigma-algebra and \(\mu=\delta_2\). Let \(f(x)=x^2-3\) and \(g(x)=100\) for every \(x\). For \(1\le p<\infty\), calculate their \(p\)-th power integrals.
The measure is concentrated at \(2\). At that point, \(f(2)=2^2-3=1\) and \(g(2)=100\). Therefore $$ \int_{\mathbb{R}}|f|^p\,d\delta_2=|1|^p=1, \qquad \int_{\mathbb{R}}|g|^p\,d\delta_2=|100|^p=100^p. $$
Both functions belong to \(L^p(\mathbb{R},\delta_2)\) for every finite \(p\ge1\), despite having quite different values away from \(2\). Check: at \(p=2\), the integrals are \(1\) and \(100^2=10{,}000\), respectively, exactly the squares of their values at the point charged by the measure.
A Common Miscalculation
A frequent mistake is to treat counting measure as if it were a probability measure and average function values. For the sequence \(a_n=3^{-n}\) on \(\mathbb{N}\), one might incorrectly write \(\int|a|^p\,d\mu=1/(3^p-1)\) and then divide by the number of terms, or claim the integral is bounded by the largest term. Counting measure assigns mass one to each individual integer; it does not normalize the total mass. The correct calculation is the full series \(\sum_{n=1}^{\infty}(3^{-p})^n=1/(3^p-1)\).
Similarly, a Dirac integral is not an integral over all values of the function weighted by their sizes: only the value at the atom contributes. On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_2)\), the function \(g(x)=100\) has \(p\)-th power integral \(100^p\), not zero and not an average over \(\mathbb{R}\).
The guiding test is always the same: specify the measure space, form \(|f|^p\), and compute its integral using that measure. Lebesgue measure produces an area integral, counting measure produces a sum, and a Dirac measure reads off one point. Finiteness of the resulting value is exactly the membership condition for finite-exponent \(L^p\).
Check Your Understanding
For each question, identify the measure space and test the integral of the \(p\)-th power rather than relying only on pointwise size.
- On \([0,1]\) with Lebesgue measure, calculate \(\int |2\chi_{[0,1/4)}-1|^p\,dm\) for \(1\le p<\infty\). Is the function in \(L^p\)?
- Under counting measure on \(\mathbb{N}\), compute \(\int |2^{-n}|^p\,d\mu\) and determine for which finite \(p\ge1\) it is finite.
- On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)\), calculate the \(p\)-th power integral of \(f(x)=x+4\). Explain why values away from \(0\) do not affect it.
- Give a bounded measurable function on an infinite-measure space that is not in \(L^p\), and verify the integral directly.
- State the precise finiteness condition that defines \(L^p(X,\mu)\) for \(1\le p<\infty\).