The space $L^2(X,\mathcal F,\nu)$ consists of equivalence classes, modulo equality almost everywhere, of measurable real functions satisfying
$$
\int_X|f|^2\,d\nu<\infty.
$$
It has inner product and induced norm
$$
\langle f,g\rangle=\int_Xfg\,d\nu,
\qquad
\|f\|_2=\left(\int_X|f|^2\,d\nu\right)^{1/2}.
$$
The <Cauchy-Schwarz inequality> makes the inner product finite and verifies the norm axioms.

To prove completeness, let $(f_n)$ be Cauchy and choose a subsequence such that
$$
\|f_{n_{k+1}}-f_{n_k}\|_2\le2^{-k}.
$$
By the triangle inequality, the partial sums
$$
g_N=\sum_{k=1}^N|f_{n_{k+1}}-f_{n_k}|
$$
have uniformly bounded $L^2$ norm. The <monotone convergence theorem> shows that their pointwise limit $g$ belongs to $L^2$ and is finite almost everywhere. The telescoping series therefore converges almost everywhere to a measurable function $f$, and
$$
\|f-f_{n_m}\|_2
\le\sum_{k=m}^{\infty}2^{-k}\longrightarrow0.
$$
The original Cauchy sequence then converges to $f$ in $L^2$. Thus the result that an <L2 space is a Hilbert space> gives
$$
\boxed{L^2(X,\mathcal F,\nu)\text{ is a Hilbert space}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 27K / a / Solution

{c}
{title2=$L^2(X,\mathcal F,\nu)$}

= L2 space
{c}
{synonym}

The space $L^2(X,\mathcal F,\nu)$ consists of almost-everywhere equivalence classes of measurable functions with $\int|f|^2\,d\nu<\infty$. The inner product
$$
\langle f,g\rangle=\int f\overline g\,d\nu
$$
induces its norm, and the <Riesz-Fischer theorem> makes it complete; hence it is a <Hilbert space>.

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 Incoming links: L2 space is a Hilbert space