Monday, 30 January 2017

real analysis - About an exercise in Rudin's book

In the book of Walter Rudin Real And Complex Analysis page 31, exercise number 10 said:
Suppose μ(X)<, {fn} a sequence of bounded complexes measurables functions on X , and fnf uniformly on X. Prove that lim

And the answer is the following



Let \epsilon > 0. Since f_n \rightarrow f uniformly, therefore there exists n_0 \in N
such that
|f_n (x) − f (x)| < \epsilon \quad ∀ n > n_0
Therefore |f (x)| < |f_{n_0} (x)| + \epsilon. Also |f_n (x)| < |f (x)| + \epsilon. Combining both
equations, we get
|f_n (x)| < |f_{n_0}| + 2\epsilon \quad ∀ n > n_0
Define g(x) = max(|f_1 (x)|, · · · , |f_{n_0 −1} (x)|, |f_{n_0} (x)| + 2\epsilon), then f_n (x) \leq g(x)
for all n. Also g is bounded. Since \mu(X)< \infty, therefore g \in \mathcal{L}^1(\mu). Now

apply DCT to get
\lim_{n \rightarrow \infty} \int_X f_n d \mu = \int_X f d \mu



What didn't I understand is why the condition f_n \rightarrow f uniformly on X is necessary? I proceed like the following



For every x \in X we have
|f_n(x)| \leq h(x)= \max_i (|f_i(x)|)
such as every f_i is bounded so h is, and in the other hand we have \mu(X) < \infty, therefore h \in \mathcal{L}^1(\mu), then we can apply DCT.



Where I am wrong please? are there any counter examples?

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