State the Dominated Convergence Theorem for Lebesgue integrals.
Let \(\{f_n\}\) be a sequence of measurable functions on a Lebesgue measurable set \(E\) which converges in measure to a function \(f\) on \(E\). Suppose that for every \(n\), \(|f_n| \leq g\) with \(g\) integrable on \(E\). Using the above theorem show that \[\begin{align*} \int_E |f_n-f| \longrightarrow 0 \, . \end{align*}\]
Solution
1. Statement of the Dominated Convergence Theorem.
1.1. Let \(\{f_n\}\) be measurable on a Lebesgue measurable set \(E\), converging almost everywhere to \(f\), with \(|f_n| \leq g\) a.e. for all \(n\), where \(g \geq 0\) is integrable on \(E\). Then \(f\) is integrable, \(|f_n-f|\to0\) in \(L^1(E)\), and in particular \(\int_E f_n\to\int_E f\).
This is Lebesgue’s dominated convergence theorem. Part (2) replaces almost-everywhere convergence by convergence in measure through a subsequence argument.
2. Proof of part (2): convergence in measure plus domination forces \(\int_E |f_n - f| \to 0\).
2.1. Some subsequence \(f_{n_k} \to f\) a.e. on \(E\).
Since \(f_n \to f\) in measure, choose indices \(n_1 < n_2 < \cdots\) with \(m\theset{|f_{n_k} - f| > 2^{-k}} < 2^{-k}\). The sets \(B_k = \theset{|f_{n_k} - f| > 2^{-k}}\) satisfy \(\sum_k m(B_k) < \infty\), so Borel–Cantelli gives \(m(\limsup_k B_k) = 0\); outside \(\limsup_k B_k\), \(f_{n_k}(x) \to f(x)\).
2.2. \(|f| \leq g\) a.e., so \(f\) is integrable and \(|f_{n_k} - f| \leq 2g\) a.e.
Pass to the a.e. limit in \(|f_{n_k}| \leq g\) using 1; then \(|f_{n_k} - f| \leq |f_{n_k}| + |f| \leq 2g\) a.e., and \(2g\) is integrable.
2.3. \(\int_E |f_{n_k} - f| \to 0\).
Apply the dominated convergence theorem to \(h_k := |f_{n_k} - f|\): by 1 and 2, \(h_k \to 0\) a.e. with \(|h_k| \leq 2g\) integrable; the theorem’s conclusion gives \(\int_E h_k \to 0\).
2.4. The full sequence satisfies \(\int_E |f_n - f| \to 0\).
If not, some \(\varepsilon > 0\) and subsequence have \(\int_E |f_{n_j} - f| \geq \varepsilon\). Applying 1 to that subsequence yields a further subsequence converging a.e., and 3 gives \(\int |f_{n_{j_l}} - f| \to 0\), contradicting \(\geq \varepsilon\). Hence the whole sequence converges to \(0\).
2.5. Q.E.D.
4 is the desired conclusion of part (2).