The main convergence theorems explain when limits can pass through integrals. They are the daily tools of analysis, probability, and asymptotic statistics.
Monotone convergence
Monotone convergence is built into the definition of the integral.
Theorem. If \(0\le f_n\uparrow f\), then \(\int f_n\,\mathrm{d}\mu\uparrow\int f\,\mathrm{d}\mu\).
Proof. The inequality \(\lim\int f_n\le\int f\) follows from monotonicity. For a simple \(s\le f\) and \(0\lt c\lt 1\), the sets \(\{f_n\ge cs\}\) increase to \(X\) on the support of \(s\). Hence \(\lim\int f_n\ge c\int s\). Let \(c\uparrow1\) and take the supremum over simple \(s\le f\).
Corollary (continuity under increasing domains). If \(A_n\uparrow A\) and \(f\ge0\), then \[ \int_{A_n}f\,\mathrm{d}\mu\uparrow\int_Af\,\mathrm{d}\mu. \] If \(f\in L^1\) and \(A_n\downarrow A\), then \(\int_{A_n}f\to\int_Af\).
Proof. Apply monotone convergence to \(f\mathbf1_{A_n}\uparrow f\mathbf1_A\). For decreasing sets, \(|f|\mathbf1_{A_n}\le|f|\) and the indicators converge pointwise, so dominated convergence applies to positive and negative parts.
Fatou's lemma
Fatou gives a one-sided lower semicontinuity of the integral.
Theorem. If \(f_n\ge0\), then \(\int\liminf_n f_n\,\mathrm{d}\mu\le\liminf_n\int f_n\,\mathrm{d}\mu\).
Proof. Let \(g_k=\inf_{n\ge k}f_n\). Then \(g_k\uparrow\liminf f_n\). By monotone convergence, \(\int g_k\to\int\liminf f_n\). Since \(g_k\le f_n\) for all \(n\ge k\), \(\int g_k\le\inf_{n\ge k}\int f_n\). Let \(k\to\infty\).
Dominated convergence
Domination gives two-sided control around pointwise convergence.
Theorem. If \(f_n\to f\) almost everywhere and \(|f_n|\le g\in L^1\), then \(f\in L^1\) and \(\int |f_n-f|\,\mathrm{d}\mu\to0\). In particular, \(\int f_n\to\int f\).
Proof. Since \(|f|\le g\) almost everywhere, \(f\in L^1\). Apply Fatou's lemma to \(2g-|f_n-f|\ge0\). This gives \(2\int g\le\liminf(2\int g-\int|f_n-f|)\), hence \(\limsup\int|f_n-f|\le0\).
Example (pointwise convergence alone is insufficient). On \((0,1)\), let \[ f_n(x)=n\mathbf1_{(0,1/n)}(x). \] Then \(f_n(x)\to0\) for every \(x>0\), but \(\int_0^1f_n\,\mathrm{d}x=1\). The mass concentrates on a shrinking set; no integrable dominating function exists. This is precisely the phenomenon controlled by uniform integrability.
Theorem (bounded convergence). If \(\mu(X)<\infty\), \(f_n\to f\) almost everywhere, and \(|f_n|\le M<\infty\), then \[ \int|f_n-f|\,\mathrm{d}\mu\longrightarrow0. \]
Proof. The constant function \(2M\) is integrable on a finite-measure space and dominates \(|f_n-f|\). Apply dominated convergence.
Scheffe-type consequence
For nonnegative functions with matching masses, pointwise convergence implies \(L^1\) convergence.
Proposition. If \(f_n,f\ge0\), \(f_n\to f\) almost everywhere, and \(\int f_n\to\int f\lt \infty\), then \(\int|f_n-f|\,\mathrm{d}\mu\to0\).
Proof. Since \(0\le f_n\wedge f\le f\) and \(f_n\wedge f\to f\) almost everywhere, dominated convergence gives \[ \int(f_n\wedge f)\longrightarrow\int f. \] Now use \[ |f_n-f|=f_n+f-2(f_n\wedge f) \] and the assumed convergence of total masses.
From domination to uniform integrability
A fixed envelope is sufficient but not necessary; what the proof really needs is uniform control of integral tails.
Remark. If \(|f_n|\le g\in L^1\), then \[ \sup_n\int_{\{|f_n|>K\}}|f_n|\,\mathrm{d}\mu \le\int_{\{g>K\}}g\,\mathrm{d}\mu\longrightarrow0. \] This is uniform integrability. Together with convergence in measure it is exactly the hypothesis of the Vitali convergence theorem.