Precompactness of subset of function space with pointwise convergence topology I need help with proving following statement. Let $M[a, b]$ - space of real (complex) value functions with topology of pointwise convergence.  Then $$A \subseteq  M[a, b] \,\text{ is precompact } \iff  \exists g \in M[a,b], |f(t)| \leq g(t)  ,\forall t \ \in \ [a,b], \ f \in \ A$$
I know, that if functions are continuous equicontinuity and uniform boundness are needed for precompactness, but don't really know what to do in this case.
 A: Assume that $\exists g \in M[a,b]$ such that $|f| \le g$ for all $f \in A$. For $x \in [a,b]$ let $\pi_x : [a,b] \to \Bbb{C}$ be the evaluation $f \mapsto f(x)$.
For every $x \in [a,b]$ we have that $\pi_x(A)$ is contained in the closed ball $\overline{K}(0,g(x))$ of radius $g(x)$ around the origin in $\Bbb{C}$.
Taking the closure yields $\overline{\pi_x(A)} \subseteq \overline{K}(0,g(x))$ and hence $\overline{\pi_x(A)}$ is compact in $\Bbb{C}$. Now we have
$$\overline{A} \subseteq \prod_{x \in [a,b]} \pi_x(\overline{A}) \subseteq  \prod_{x \in [a,b]} \overline{\pi_x(A)}$$
and the latter is a product of compact sets and hence compact. Therefore $\overline{A}$ is also compact.
Conversely, assume that $A$ is precompact. Then for every $x \in X$ since $\pi_x$ is continuous, we have that $\pi_x(\overline{A})$ is compact in $\Bbb{C}$ as an image of a compact set. In particular, $\pi_x(\overline{A})$ is bounded and then from
$$\pi_x(A) \subseteq \pi_x(\overline{A})$$
we see that is  $\pi_x(A)$ is also bounded. Hence there exists a positive real number which we can denote by $g(x)$ such that $\pi_x(A) \subseteq \overline{K}(0,g(x))$.
Hence for any $f \in A$ and $x \in [a,b]$ we have
$$f(x) = \pi_x(f) \in \pi_x(A) \subseteq \overline{K}(0,g(x)) \implies |f(x)| \le g(x).$$
