Convergence in Norm Implies Convergence in Measure

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \R, p \ge 1$.

Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of $p$-integrable functions.

Also, let $f: X \to \R$ be a $p$-integrable function.

Suppose that $f_n$ converges in norm to $f$ (in the $L^p$-norm).

Then $f_n$ converges in measure to $f$ (in $\mu$).

That is:

$\ds \operatorname {\LL^{\textit p}-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f \implies \operatorname {\mu-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f$

Proof

Let $\epsilon > 0$.

Then:

\(\ds \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} \ge \epsilon } }\)

\(=\)

\(\ds \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x}^p \ge \epsilon^p } }\)

\(\ds \)

\(\le\)

\(\ds \frac{1}{\epsilon^p} \int \size {f_n - f}^p \rd \mu\)

Markov's Inequality

\(\ds \)

\(\to\)

\(\ds 0\)

as $n \to +\infty$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $16.4 \ \text{(i)}$