Convergence in $L^p$

A sequence $(f_n)$ in a $L^p$ space $L^p(X,\mathcal A,\mu)$ converges in $L^p$ to $f\in L^p(X,\mathcal A,\mu)$ if

\|f_n-f\|_p \to 0 \quad \text{as } n\to\infty,

where $(X,\mathcal A,\mu)$ is a measure space. For $p=\infty$ , the norm is the essential supremum norm.

Relation to convergence in measure

For $1\le p<\infty$ and $\varepsilon>0$ , Markov's inequality gives

\mu(\{|f_n-f|>\varepsilon\}) \le \varepsilon^{-p}\|f_n-f\|_p^p

for every $n$ . Consequently, convergence in $L^p$ implies convergence in measure.

Examples

On $([0,1],\mathcal{B},\lambda)$ , the functions $f_n(x)=x^n$ satisfy $f_n\to 0$ in $L^p$ for every $1\le p<\infty$ , since $\|f_n\|_p^p=\int_0^1 x^{np}\,dx=\frac{1}{np+1}\to 0.$
On the same space, for fixed $1\le p<\infty$ , the functions $g_n=n^{1/p}\mathbf{1}_{[0,1/n]}$ converge to $0$ in measure but not in $L^p$ , because $\|g_n\|_p^p = \int_0^{1/n} n\,dx = 1$ for every $n$ .