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The Real And Complex Number Systems

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which implies that<br />

∫ b m∑<br />

a<br />

k=1<br />

|f nk − g| 2 dx <<br />

which implies that, by Fatou’s lemma,<br />

m∑<br />

k=1<br />

1<br />

2 k<br />

∫ b<br />

a<br />

lim inf ∑ m ∫ b m∑<br />

|f nk − g| 2 dx ≤ lim inf |f nk − g| 2 dx<br />

m→∞ m→∞<br />

k=1<br />

a<br />

k=1<br />

∞∑<br />

∫ b<br />

= |f nk − g| 2 dx < 1.<br />

k=1<br />

a<br />

That is,<br />

which implies that<br />

∫ b ∞∑<br />

a<br />

k=1<br />

|f nk − g| 2 dx < 1<br />

∞∑<br />

|f nk − g| 2 < ∞ a.e. on [a, b]<br />

k=1<br />

which implies that f nk → g a.e. on [a, b] .<br />

Note: <strong>The</strong> reader can see the book, Measure and Integral (An Introduction<br />

to <strong>Real</strong> Analysis) written by Richard L. Wheeden and<br />

Antoni Zygmund, pp 75.<br />

(4) <strong>The</strong>re is another proof by using Egorov’s <strong>The</strong>orem: Let {f k } be a<br />

measurable functions defined on a finite measurable set E with finite limit<br />

function f. <strong>The</strong>n given ε > 0, there exists a closed set F (⊆ E) , where<br />

|E − F | < ε such that<br />

f k → f uniformly on F.<br />

Proof: If f ≠ g on [a, b] , then h := |f − g| ≠ 0 on [a, b] . By continuity<br />

of h, there exists a compact subinterval [c, d] such that |f − g| ≠ 0. So, there<br />

exists m > 0 such that h = |f − g| ≥ m > 0 on [c, d] . Since<br />

∫ b<br />

a<br />

|f n − g| 2 dx → 0 as n → ∞,<br />

26

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