By Nicolas Lerner

This textbook presents a close remedy of summary integration thought, building of the Lebesgue degree through the Riesz-Markov Theorem and in addition through the Carathéodory Theorem. additionally it is a few common homes of Hausdorff measures in addition to the elemental homes of areas of integrable features and conventional theorems on integrals counting on a parameter. Integration on a product area, swap of variables formulation in addition to the development and research of classical Cantor units are taken care of intimately. Classical convolution inequalities, comparable to Young's inequality and Hardy-Littlewood-Sobolev inequality are confirmed. The Radon-Nikodym theorem, notions of harmonic research, classical inequalities and interpolation theorems, together with Marcinkiewicz's theorem, the definition of Lebesgue issues and Lebesgue differentiation theorem are additional issues incorporated. a close appendix presents the reader with a variety of parts of effortless arithmetic, similar to a dialogue round the calculation of antiderivatives or the Gamma functionality. The appendix additionally presents extra complex fabric equivalent to a few uncomplicated homes of cardinals and ordinals that are worthwhile within the research of measurability.

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**Extra resources for A Course on Integration Theory: including more than 150 exercises with detailed answers**

**Example text**

5. Show that R is equipotent to P(N) (hint: use dyadic expansions). Show that R is not countable. Answer. 2. 19) is bijective from R onto (−1, 1), which is equipotent to (0, 1) (x → (x + 1)/2). We have seen in the previous exercise that P(N) is equipotent to {0, 1}N, the set of mappings from N into {0, 1}. We have thus to prove that {0, 1}N is equipotent to (0, 1). Let x be in (0, 1). With E standing for the ﬂoor function (see the footnote on page 16), we deﬁne for any integer k ≥ 1, xk = E(2k x) − 2E(2k−1 x) = pk (x).

Let f1 , . . , fN be measurable functions from X −→ R+ . Then f1 + · · · + fN is measurable and X (f1 + · · · + fN )dμ = X f1 dμ + · · · + X fN dμ. Proof. Using induction on N , it is enough to prove the lemma for N = 2. 3, let sk , sk be simple functions (j) 0 ≤ sk ↑ fj , j = 1, 2. 1, we get (j) sk dμ ↑ X fj dμ. 7) the result of the lemma. 4 (Fatou’s Lemma). Let (X, M, μ) be a measure space where μ is a positive measure. Let (fn )n≥0 be a sequence of measurable functions from X → R+ . The following inequality holds: (lim inf fn )dμ ≤ lim inf X n n fn dμ .

Let (X, M, μ) be a measure space where μ is a positive measure and let f : X −→ Y be a mapping. The set N = {B ⊂ Y, f −1 (B) ∈ M} is a σalgebra on Y : it is the largest σ-algebra on Y making f measurable. The so-called pushforward measure f∗ (μ) is a positive measure deﬁned on N by f∗ (μ)(B) = μ f −1 (B) . If g : Y −→ Z is another mapping, we have (g ◦ f )∗ = g∗ ◦ f∗ . Proof. 5. To check that f∗ (μ) is a positive measure deﬁned on N , we consider a sequence (Bk )k∈N of pairwise disjoint elements of N and we note that f −1 (Bk ) k∈N is a pairwise disjoint sequence of M and thus f∗ (μ) ∪k∈N Bk = μ f −1 (∪k∈N Bk ) = μ ∪k∈N f −1 (Bk ) = k μ f −1 (Bk ) = k f∗ (μ)(Bk ).