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Question 19. Suppose \({f_n} \subset L^1(\mu)\) and \(f_n \rightarrow f\) uniformly.

1. If \(\mu(x) < \infty\) then \(f \in L^1(\mu)\) and \(\int f_n \rightarrow \int f\)
2. if \(\mu(x) = \infty\), then the previous part can fail.

Proof:(part a) Suppose \({f_n} \subset L^1(\mu)\), \(f_n \rightarrow f\) uniformly and \(\mu(x) < \infty\). So then there exists some \(N \in \mathbb{N}\) such that \(|f_n(x) - f(x)| \leq 1\) for all \(n \geq N\) and for all \(x\in X\).

This means that \(\int|f| \leq \int|f-f_N| + \int|f_N| \leq \int 1 + \int |f_N| = \mu(x) + \int |f_N|\)

Note that \(\mu(x) + \int|f_N| < \infty\). So \(f \in L^1(\mu)\)

Now, let \(g= 1 +|f|\). Then \(g \geq 0\) and \(\int g = \mu(x) + \int|f| < \infty\). So \(g \in L^1(\mu)\) and \(|f_n| \leq g\) for all \(n \geq N\). So the dominated convergence theorem implies that \(lim_(n \rightarrow \infty) \int f_n = \int f\).

(part b) Let \(X = \mathbb{R}\) and \(f_n(x) = \begin{cases} 0 \text{ if } x\leq n \\ \frac{1}{x} \text{ if } x>n \end{cases} \)

It is well known that \(\frac{1}{x}\) converges to \(0\) as \(x\rightarrow \infty\), so naturally \(f_n(x) \rightarrow 0\) uniformly as \(n\rightarrow \infty\). Further, we know \(\int f_n(x) = \int _n ^\infty \frac{1}{x} = \infty\). But \(\int 0 = 0 \). So the previous part fails since we didn't have the condition that \(\mu(x) <\infty\)

Question 20. A generalized form of the Dominated Convergence Theorem: