Ph.D. Qualifying Exam – Spring 2004

Sixth problemLet (X, B, µ) be a finite measure space. Suppose that (f n ) n∈N is a sequence of functionsin L 1 (µ), converging almost everywhere to an L 1 (µ) function f. Suppose also that‖f n ‖ 1 −−−−→ ‖f‖ 1n→∞∫1. Prove that for every measurable set A, |f n | dµ −−−−→n→∞2. Prove that ‖f n − f‖ 1 −−−−→n→∞ 0. SolutionA∫A|f|.1 For every integer n, defineg n = Min (|f n |, |f|) = |f n| + |f| − ∣ |fn | − |f| ∣ 2Since (f n ) n∈N converges almost everywhere to f, it follows that (g n ) n∈N converges almosteverywhere to |f| and is dominated by |f|.Let A be any measurable set. Then (g n 1 A ) n∈N converges almost everywhere to |f|1 Aand is dominated by |f|1 A . The dominated convergence theorem implies that∫ ∫limn→∞The same argument shows thatlimn→∞∫AXg n dµ =∫g n dµ =AX|f| dµ|f| dµNow, by definition of g n , the function |f n | − g n is nonnegative for every integer n andthus we have0 ( )|f n | − g n 1A |f n | − g n∫so that0 ∫( )|fn | − g n dµ ( )|fn | − g n dµA∫XBoth terms on the righthandside tend to |f| dµ, thusX∫(|fn | − g n ) dµ = 0So∀n ∈ N∫Alimn→∞∫|f n | dµ =AA(|fn | − g n)dµ +∫Ag n dµ −−−−→n→∞∫A|f| dµ10