Throughout, f and f_n (n\in\N) are measurable functions X\to\R. • Global convergence in measure implies local convergence in measure. The converse, however, is false;
i.e., local convergence in measure is strictly weaker than global convergence in measure, in general. • If, however, \mu (X) or, more generally, if f and all the f_n vanish outside some set of finite measure, then the distinction between local and global convergence in measure disappears. • If \mu is
σ-finite and (
fn) converges (locally or globally) to f in measure, there is a subsequence converging to f
almost everywhere. The assumption of
σ-finiteness is not necessary in the case of global convergence in measure. • If \mu is \sigma-finite, (f_n) converges to f locally in measure
if and only if every subsequence has in turn a subsequence that converges to f almost everywhere. • In particular, if (f_n) converges to f almost everywhere, then (f_n) converges to f locally in measure. The converse is false. •
Fatou's lemma and the
monotone convergence theorem hold if almost everywhere convergence is replaced by (local or global) convergence in measure. • If \mu is \sigma-finite, Lebesgue's
dominated convergence theorem also holds if almost everywhere convergence is replaced by (local or global) convergence in measure. • If X=[a,b]\subseteq\R and
μ is
Lebesgue measure, there are sequences (g_n) of step functions and (h_n) of continuous functions converging globally in measure to f. • If f and f_n are in
Lp(μ) for some p>0 and (f_n) converges to f in the p-norm, then (f_n) converges to f globally in measure. The converse is false. • If f_n converges to f in measure and g_n converges to g in measure then f_n+g_n converges to f+g in measure. Additionally, if the measure space is finite, f_n g_n also converges to fg. ==Counterexamples==