Bounded Measure functions on a set of finite measure.

Question

In Real Analysis of Royden and Fritzpatrick's book page 77, Proposition 8 proof first sentence states that " Since the convergence is uniform and each $f_n$ is bounded, the limit function f is bounded". I use the definition on uniform on E but I cannot find the $M$.I really appreciate your help if you would like to give me a detail answer of the sentence. Thank you so much.

Answer 1

Let $X \neq \varnothing$ and $(Y,d_Y)$ be a metric space and $f_n: X \to (Y,d_Y)$ be a sequence of bounded functions converging uniformly to $f$. Since each $f_n$ is bounded, there exists $b \in Y$ and a sequence of positive real numbers $(M_n)$ such that $f_n(X) \subset B_{d_Y}(b, M_n)$ for all $n\in \Bbb N$. Since $(f_n)$ converges uniformly to $f$, there exists $n_0$ such that $\sup_{x \in X} d_Y(f_{n_0}(x), f(x)) < 1$. Hence, $f(x) \in B_{d_Y}(f_{n_0}(x), 1)$ for all $x \in X$. But $B_{d_Y}(f_{n_0}(x), 1) \subset B_{d_Y}(b, M_{n_0} + 1)$ for all $x \in X$, so $f(x) \in B_{d_Y}(b, M_{n_0} + 1)$ for all $x \in X$, thus $f(X)\subset B_{d_Y}(b, M_{n_0} + 1)$, and so $f$ is bounded.