## Advanced calculus. Problems and applications to science and by Hugo. Rossi

By Hugo. Rossi

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**Example text**

4, we obtain (3). To prove (4), we deﬁne E = {f = +∞}, and we note that μ(E) > 0 implies for all integers n ≥ 1, that f dμ ≥ X entailing X f dμ ≥ n E dμ = nμ(E), E f dμ = +∞. 2. Let (X, M, μ) be a measure space where μ is a positive measure. The space L1 (μ) is deﬁned as the quotient of L1 (μ) (cf. e. (f ∼ g means μ({x ∈ X, f (x) = g(x)}) = 0). 3. e. since μ(N1 ∪ N2 ) = 0. 36 Chapter 1. 1(1). , then X gdμ. 1. 4. Let (X, M, μ) be a measure space where μ is a positive measure. (1) The mapping from L1 (μ) into C deﬁned by f → X f dμ is a linear form.

Then lim inf xn is the smallest accumulation point of the sequence and lim sup xn the largest. We have lim inf xn ≤ lim sup xn and equality holds if and only if the sequence is converging to this value. Proof. Using the homeomorphism ψ0 deﬁned above (cf. 19)) we can assume that (xn )n∈N is a sequence in [−1, 1]. , a limit point of subsequence, (xnk )k∈N , (n0 < n1 < n2 < · · · < nk < nk+1 < · · · ), then y ←− xnk ≤ sup xl −→ lim sup xn , k→+∞ l≥nk k→+∞ where the second limit comes from a subsequence of a converging sequence.

8. Let (X, M, μ) be a measure space where μ is a positive measure. Let f be in L1 (μ) and let (fn )n∈N be a sequence of functions in L1 (μ) such that the following properties hold. , (2) limn fn L1 (μ) = f L1(μ) . Then limn fn − f L1 (μ) = 0. 9. To sum-up, for a sequence (fn ) in L1 (μ), f ∈ L1 (μ), pointwise / fn convergence ⎫ ⎪ ⎬ f ⎪ ⎭ and limn fn L1 (μ) = f =⇒ fn L1 (μ) / f . 6) L1 (μ) The following proposition is an important consequence of the Lebesgue dominated convergence theorem. 10. Let (X, M, μ) be a measure space where μ is a positive measure.