Different modes of convergence encode different kinds of control: pointwise, almost everywhere, in measure, in norm, and uniformly. Knowing the implications prevents false limit interchanges.
Almost everywhere and in measure
Almost-everywhere convergence is pointwise outside a null set; convergence in measure controls exceptional sets quantitatively.
Definition. \(f_n\to f\) almost everywhere if \(\mu(\{x:f_n(x)\not\to f(x)\})=0\). It converges in measure if \(\mu(|f_n-f|>\varepsilon)\to0\) for every \(\varepsilon>0\).
Theorem. On a finite measure space, almost-everywhere convergence implies convergence in measure.
Proof. For fixed \(\varepsilon\), define \(A_n=\bigcup_{k\ge n}\{|f_k-f|>\varepsilon\}\). Then \(A_n\downarrow A\), where \(A\) is contained in the null set of nonconvergence. Since \(\mu(X)\lt \infty\), \(\mu(A_n)\downarrow0\), and \(\mu(|f_n-f|>\varepsilon)\le\mu(A_n)\).
Lp convergence
Norm convergence controls average size of errors.
Definition. For \(1\le p\lt \infty\), \(f_n\to f\) in \(L^p\) if \(\int |f_n-f|^p\,\mathrm{d}\mu\to0\). For \(p=\infty\), use essential sup norm.
Proposition. If \(f_n\to f\) in \(L^p\) for \(1\le p\lt \infty\), then \(f_n\to f\) in measure.
Proof. Markov's inequality gives \(\mu(|f_n-f|>\varepsilon)\le\varepsilon^{-p}\int|f_n-f|^p\,\mathrm{d}\mu\), which tends to zero.
Subsequence principle
Convergence in measure is almost-everywhere convergence after subselection.
Theorem. If \(f_n\to f\) in measure, then some subsequence converges to \(f\) almost everywhere.
Proof. Choose \(n_k\) with \(\mu(|f_{n_k}-f|>2^{-k})\lt 2^{-k}\). The sum of the exceptional measures is finite. By Borel-Cantelli, almost every point belongs to only finitely many exceptional sets, so \(f_{n_k}(x)\to f(x)\) almost everywhere.
Proposition. On a finite measure space, convergence in measure has an a.e.-subsequence characterization: \[ f_n\to f\text{ in measure} \quad\Longleftrightarrow\quad \text{every subsequence }(f_{n_k})\text{ has a further subsequence converging a.e. to }f. \]
Proof. The forward implication applies the preceding subsequence extraction to each subsequence, which still converges in measure. Conversely, if convergence in measure fails, there are \(\varepsilon,\delta>0\) and a subsequence with \[ \mu(|f_{n_k}-f|>\varepsilon)\ge\delta \] for every \(k\). No further subsequence can converge almost everywhere to \(f\), because on a finite measure space a.e. convergence implies convergence in measure, contradicting the fixed lower bound.
Uniform convergence
Uniform convergence is stronger than pointwise convergence and preserves continuity.
Proposition. If \(f_n\to f\) uniformly and each \(f_n\) is continuous on a metric space, then \(f\) is continuous.
Proof. Fix \(x\) and \(\varepsilon>0\). Choose \(n\) with \(\|f_n-f\|_\infty\lt \varepsilon/3\), then choose a neighborhood of \(x\) on which \(f_n\) varies by less than \(\varepsilon/3\). The triangle inequality gives continuity of \(f\).
Implication map
On a finite measure space, the valid unconditional arrows are
Counterexamples to the missing arrows.
A.e. does not imply \(L^1\): \(f_n=n\mathbf1_{(0,1/n)}\to0\) a.e. on \((0,1)\), but \(\|f_n\|_1=1\).
In measure does not imply a.e.: enumerate the dyadic intervals in \([0,1]\) level by level and let \(f_n\) be their indicators. Their lengths tend to zero, so \(f_n\to0\) in measure, but every point lies in one interval at each level, so the sequence does not converge pointwise anywhere away from dyadic endpoint conventions.
\(L^p\) does not imply uniform: \(f_n=\mathbf1_{(0,1/n)}\) has \(\|f_n\|_p=n^{-1/p}\to0\), while \(\|f_n\|_\infty=1\).
Finite measure matters: on \(\mathbb R\), \(f_n=\mathbf1_{[n,n+1]}\to0\) pointwise but not in measure.