Lebesgue integration Study Guide

📖 Core Concepts Measure space – a triple \((E,\mathcal{X},\mu)\) with a set \(E\), a σ‑algebra \(\mathcal{X}\) of measurable subsets, and a measure \(\mu\) assigning a non‑negative extended real number to each set. Measurable function – \(f:E\to\mathbb{R}\) is measurable if \(\{x\in E: f(x)>t\}\in\mathcal{X}\) for every real \(t\). Simple function – finite linear combination of indicator functions: \[ s(x)=\sum{k=1}^{n}ak\mathbf 1{Sk}(x),\qquad ak\in\mathbb{R},\; Sk\in\mathcal{X}. \] Lebesgue integral of a simple function – \(\displaystyle\intE s\,d\mu = \sum{k=1}^{n} ak\,\mu(Sk)\). General non‑negative function – defined as the supremum of integrals of increasing simple approximations: \[ \intE f\,d\mu = \supn \intE sn\,d\mu,\qquad sn\uparrow f. \] Signed/complex functions – write \(f=f^+-f^-\) (or \(h=u+iv\)); integrate each part separately. Null set & “almost everywhere’’ (a.e.) – a set \(N\) with \(\mu(N)=0\); two functions equal a.e. differ only on a null set. Fundamental theorems – Linearity, Monotonicity, Monotone Convergence (MCT), Fatou’s Lemma, Dominated Convergence (DCT). --- 📌 Must Remember Integral of a simple function: \(\displaystyle\int s\,d\mu=\sum ak\mu(Sk)\). Lebesgue integral of \(f\ge0\): supremum of simple‑function integrals. MCT: If \(fk\uparrow f\) (non‑negative), then \(\displaystyle\int f = \lim{k\to\infty}\int fk\). Fatou: \(\displaystyle \int \liminf fk \le \liminf \int fk\) for \(fk\ge0\). DCT: If \(fk\to f\) a.e. and \(|fk|\le g\) with \(\int g<\infty\), then \(\displaystyle\int f = \lim{k\to\infty}\int fk\). Linearity: \(\int (a f + b g)=a\int f + b\int g\). Monotonicity: \(f\le g\) a.e. \(\Rightarrow\) \(\int f\le\int g\). Dirichlet function on \([0,1]\): Lebesgue integral = 0 because rationals form a null set. --- 🔄 Key Processes Constructing simple approximations for a non‑negative measurable \(f\): Partition the range into intervals \([k/2^n,(k+1)/2^n)\). Define \(sn(x)=\frac{k}{2^n}\) on the set where \(f(x)\) falls in that interval. Obtain an increasing sequence \(sn\uparrow f\). Integrating a signed function \(f\): Compute \(f^+=\max\{f,0\}\) and \(f^-=\max\{-f,0\}\). Evaluate \(\int f^+\) and \(\int f^-\) (both non‑negative). If at least one is finite, set \(\displaystyle\int f = \int f^+ - \int f^-\). Applying DCT: Verify pointwise a.e. convergence \(fk\to f\). Find an integrable dominating function \(g\) with \(|fk|\le g\). Conclude \(\displaystyle\int f = \lim\int fk\). --- 🔍 Key Comparisons Riemann vs. Lebesgue (integration) Domain partition: Riemann → subintervals (vertical rectangles); Lebesgue → range intervals (horizontal slabs). Functions handled: Riemann fails for highly discontinuous functions (e.g., Dirichlet); Lebesgue integrates them (often giving 0). Limit‑integral interchange: Riemann needs strong uniform/continuous hypotheses; Lebesgue provides MCT, DCT, Fatou. Simple function vs. General measurable function Simple: finite sum of weighted indicator sets, integral is a finite sum. General non‑negative: limit of an increasing simple sequence; integral defined as supremum of simple integrals. --- ⚠️ Common Misunderstandings “If the Lebesgue integral exists, the Riemann integral must exist.” – False; many Lebesgue‑integrable functions (Dirichlet) have no Riemann integral. “A null set can be ignored completely.” – You may modify a function on a null set without changing its integral, but null sets matter when proving a.e. statements. “Dominated convergence works without a dominating function.” – DCT requires an integrable \(g\) with \(|fk|\le g\). --- 🧠 Mental Models / Intuition Horizontal‑slab picture: Imagine slicing the graph horizontally; each slice’s “width’’ is the measure of the set where the function’s value lies in that slice. Summing width × height (the slice thickness) yields the Lebesgue integral. Simple functions as “building blocks”: Think of Lego bricks—each brick (indicator of a measurable set) has a known “volume” (\(ak\mu(Sk)\)). Assemble them to approximate any shape (function). --- 🚩 Exceptions & Edge Cases Integrability of signed functions: If both \(\int f^+\) and \(\int f^-\) are infinite, the difference is undefined – the function is not Lebesgue integrable. Complex‑valued integrals: Must check integrability of both real and imaginary parts; otherwise the integral does not exist. MCT fails for non‑monotone sequences – use Fatou or DCT instead. --- 📍 When to Use Which Use simple‑function approximation when you need to define the integral of a new non‑negative measurable function. Apply MCT when you have a monotone increasing sequence of non‑negative functions (e.g., partial sums of a series of non‑negative terms). Apply Fatou’s Lemma to obtain a lower bound when only pointwise \(\liminf\) information is available. Apply DCT when the sequence converges pointwise (or a.e.) and you can exhibit an integrable dominating function (common in Fourier analysis, probability). Use linearity & monotonicity for quick algebraic manipulations and inequality checks. --- 👀 Patterns to Recognize “Increasing simple sequence → supremum of integrals” → whenever a problem asks for \(\int f\) with \(f\ge0\), look for a construction of \(sn\). Null‑set appearances → if a function differs from a simpler one only on a set of measure 0, they have the same integral. Dominating function of the form \(g(x)=C\cdot h(x)\) where \(h\) is known integrable → a typical DCT cue. Series of non‑negative terms → suspect MCT can be invoked to swap sum and integral. --- 🗂️ Exam Traps Choosing “Riemann” instead of “Lebesgue” for a function with countably many discontinuities (e.g., Dirichlet). The correct answer will emphasize the null‑set argument. Confusing “finite” simple sum with “infinite” integral – forgetting that if any coefficient \(ak\) is infinite, the simple integral can be \(+\infty\). Misapplying DCT without a dominating integrable function – answer choices may present a bounded sequence but no integrable bound; the correct choice will note the missing domination. Assuming Fatou gives equality – it only provides an inequality; answer options that claim \(\int \liminf fk = \liminf \int fk\) are traps. Neglecting the “almost everywhere’’ qualifier – statements about pointwise inequality must be qualified a.e.; an option that omits “a.e.” is usually wrong.