What It Means to Converge in \(L^p\)
Completeness of \(L^p\) says that a Cauchy sequence has a limit in the space. We now describe what it means for a sequence to approach that limit and identify consequences of this mode of convergence. The central quantity is not the value of the functions at each point, but the \(L^p\) norm of their difference.
As in earlier tutorials, \(L^p(X,\mu)\) consists of equivalence classes of measurable functions, where functions equal almost everywhere represent the same element. Thus convergence in \(L^p\) is a statement about these equivalence classes. Any measurable representatives may be used when writing integrals or level sets; changing a representative on a null set does not affect the resulting norm or measure.
For a finite exponent, the integral condition follows directly from the definition of the norm: \(\|f_n-f\|_p^p=\int_X|f_n-f|^p\,d\mu\). Since taking the \(p\)th power is continuous on the nonnegative real numbers, the norm tends to zero exactly when this integral does. Convergence in \(L^p\) is therefore a measure of the total \(p\)th-power error across the space, rather than a requirement that the error be small at every point.
A Basic Consequence: Convergence in Measure
Small \(L^p\) error controls the measure of the set where the error exceeds a chosen tolerance. This gives a useful consequence on any measure space, whether or not the whole space has finite measure.
Proof. Since \(f_n,f\in L^p\), linear closure of \(L^p\) implies \(f_n-f\in L^p\). Apply the level-set estimate for \(L^p\) to \(f_n-f\) at threshold \(\varepsilon\). For every \(n\), $$ \mu\bigl(\{|f_n-f|>\varepsilon\}\bigr) \leq \frac{\|f_n-f\|_p^p}{\varepsilon^p}. $$ The denominator is a fixed positive number, while \(\|f_n-f\|_p\to0\). Consequently the right-hand side tends to zero, proving the claim.
The conclusion is convergence in measure: for each fixed tolerance, the measure of the set on which the error is larger than that tolerance tends to zero. The theorem does not require \(\mu(X)\) to be finite. It also does not say that the error is small everywhere, or that the functions converge at every point. Those are distinct kinds of statements.
Norms and Moments Along a Convergent Sequence
Another immediate-looking consequence is important in applications: if functions converge in \(L^p\), their \(L^p\) norms converge as real numbers. For finite \(p\), this also means their \(p\)th-power integrals converge. The Reverse Triangle Inequality gives the needed estimate.
Proof. The Reverse Triangle Inequality states that $$ \bigl|\|f_n\|_p-\|f\|_p\bigr|\leq\|f_n-f\|_p. $$ The right-hand side tends to zero, so the norms converge. This argument applies to \(p=\infty\) as well. For finite \(p\), the function \(t\mapsto t^p\) is continuous on \([0,\infty)\). Taking \(p\)th powers of the convergent nonnegative numbers \(\|f_n\|_p\) therefore gives $$ \int_X|f_n|^p\,d\mu =\|f_n\|_p^p \longrightarrow \|f\|_p^p =\int_X|f|^p\,d\mu. $$ This proves both claims.
Worked Examples
Worked Example: A Scalar Sequence on a Fixed Set
Let \(E\in\mathcal{F}\) satisfy \(0<\mu(E)<\infty\), fix \(1\leq p<\infty\), and consider \(f_n=c_n\mathbf{1}_E\) and \(f=c\mathbf{1}_E\), where \(c_n,c\) are real numbers. Their difference is $$ f_n-f=(c_n-c)\mathbf{1}_E, $$ so direct calculation gives $$ \|f_n-f\|_p^p =\int_X|c_n-c|^p\mathbf{1}_E\,d\mu =|c_n-c|^p\mu(E). $$ Taking the \(p\)th root, $$ \|f_n-f\|_p=|c_n-c|\mu(E)^{1/p}. $$ Because \(0<\mu(E)<\infty\), this norm tends to zero exactly when \(c_n\to c\). For example, if \(c_n=2+1/n\) and \(c=2\), then \(\|f_n-f\|_p=\mu(E)^{1/p}/n\to0\). The measure and exponent determine the scale of the error, while convergence is determined by the scalar coefficients.
Worked Example: Shrinking Supports in Finite \(L^p\)
On \((0,1)\) with Lebesgue measure, let \(f_n=\mathbf{1}_{(0,1/n)}\) and let \(f=0\). For every finite \(p\geq1\), $$ \|f_n-f\|_p^p =\int_0^1\mathbf{1}_{(0,1/n)}(x)\,dx =\frac1n, $$ and hence $$ \|f_n-f\|_p=n^{-1/p}\longrightarrow0. $$ Thus \(f_n\to0\) in every finite \(L^p(0,1)\). But \(\|f_n\|_\infty=1\) for every \(n\), since the interval \((0,1/n)\) has positive measure. The sequence does not converge to zero in \(L^\infty\). A small support makes the finite-\(p\) error small even though the error retains height one; the \(L^\infty\) norm records that height and does not become small.
Worked Example: Convergence in \(L^p\) Without Pointwise Convergence
Let \(X=[0,1)\) with Lebesgue measure. For each positive integer \(m\), partition \(X\) into the half-open intervals $$ I_{m,j}=[j2^{-m},(j+1)2^{-m}),\qquad j=0,1,\ldots,2^m-1. $$ Form a single sequence by listing, in order of increasing \(m\), the functions \(\mathbf{1}_{I_{m,0}},\ldots,\mathbf{1}_{I_{m,2^m-1}}\) for each level \(m\). Every function at level \(m\) has norm $$ \|\mathbf{1}_{I_{m,j}}\|_p =\left(\int_X\mathbf{1}_{I_{m,j}}\,d\mu\right)^{1/p} =2^{-m/p}. $$ As the sequence progresses, its level \(m\) tends to infinity; therefore these norms tend to zero, and the sequence converges to zero in \(L^p(X)\) for every finite \(p\geq1\).
However, fix any \(x\in[0,1)\). At each level \(m\), exactly one interval in the partition contains \(x\). The corresponding indicator has value \(1\) at \(x\), so the sequence has value \(1\) infinitely often. At every level \(m\geq2\), there are also intervals not containing \(x\), so the sequence has value \(0\) infinitely often. The values at \(x\) consequently do not converge. This example shows that convergence in \(L^p\) alone does not guarantee pointwise convergence of the full sequence.
What to Keep in Mind
Finite-\(p\) convergence controls an averaged error: the integral of the \(p\)th power of the difference tends to zero. It follows that the functions converge in measure and that their norms and \(p\)th moments converge. But the size of an average does not rule out errors of fixed height on sets whose measures shrink, as the indicator example demonstrates. Nor does it prevent exceptional sets from moving through the space, as in the partition example.
The exponent matters. For finite \(p\), a function can have a large value on a sufficiently small set and still have a small \(L^p\) norm. In \(L^\infty\), the essential supremum measures the size of the error outside null sets rather than averaging it over the space. Thus convergence in a finite \(L^p\) norm and convergence in \(L^\infty\) are genuinely different requirements.
Check Your Understanding
Use the definition and the proved consequences to answer the following questions.
- For \(1\leq p<\infty\), how is \(\|f_n-f\|_p\) related to the integral of \(|f_n-f|^p\)?
- Why does the level-set estimate show convergence in measure without requiring \(\mu(X)<\infty\)?
- If \(f_n\to f\) in \(L^p\), what does the Reverse Triangle Inequality imply about \(\|f_n\|_p\)?
- Why do the shrinking indicator functions converge to zero in finite \(L^p(0,1)\) but not in \(L^\infty(0,1)\)?
- In the partition example, why do the function values at each fixed point fail to converge even though the sequence converges in \(L^p\)?