Completeness of Lᵖ: The Riesz–Fischer Theorem

What completeness means here

A sequence (fₙ) in Lᵖ is Cauchy in norm if ‖fₙ − fₘ‖ₚ → 0 as n, m → ∞. The Riesz–Fischer theorem says every such sequence has a limit f ∈ Lᵖ with ‖fₙ − f‖ₚ → 0. In other words Lᵖ is a complete normed space — a Banach space. Without this, limits of approximations could leak out of the space and analysis in Lᵖ would collapse.

The proof, step by step

1. Extract a rapid subsequence. From the Cauchy sequence pick n₁ < n₂ < … so that ‖f_{n_{k+1}} − f_{n_k}‖ₚ ≤ 2^(−k). Write g_k = f_{n_{k+1}} − f_{n_k}, so Σ ‖g_k‖ₚ ≤ Σ 2^(−k) = 1 is finite.
2. Build a dominating function. Let G = |f_{n_1}| + Σ |g_k|. By Minkowski and monotone convergence, ‖G‖ₚ ≤ ‖f_{n_1}‖ₚ + Σ ‖g_k‖ₚ < ∞, so G ∈ Lᵖ and in particular G(x) < ∞ almost everywhere.
3. Get a pointwise limit. Where G(x) < ∞, the series f_{n_1}(x) + Σ g_k(x) converges absolutely, and its partial sums telescope to f_{n_k}(x). Call the limit f(x). So f_{n_k} → f almost everywhere, and |f| ≤ G, hence f ∈ Lᵖ.
4. Upgrade to norm convergence. Since |f_{n_k} − f|ᵖ ≤ (2G)ᵖ ∈ L¹ and f_{n_k} → f a.e., the dominated convergence theorem gives ‖f_{n_k} − f‖ₚ → 0. Finally a Cauchy sequence whose subsequence converges, converges, so ‖fₙ − f‖ₚ → 0.

Why the subsequence pins down the whole sequence.

Given ε > 0, Cauchy gives N with ‖fₙ − fₘ‖ₚ < ε/2 for n, m ≥ N.
We also have ‖f_{n_k} − f‖ₚ → 0, so pick k with n_k ≥ N and
        ‖f_{n_k} − f‖ₚ < ε/2.
Then for every n ≥ N, by Minkowski (the triangle inequality):
        ‖fₙ − f‖ₚ ≤ ‖fₙ − f_{n_k}‖ₚ + ‖f_{n_k} − f‖ₚ
                  <  ε/2 + ε/2 = ε.
Hence ‖fₙ − f‖ₚ → 0. ∎