monotone convergence without 1 point Suppose we have a sequence of nonnegative continuous functions $\{f_i(x)\}_1^\infty$ such that $\sum_{i=1}^\infty f_i(x)$ converges for any $x\ne 0$ and $F(x)$ is some continuous function that satisfies $F(x)=\sum_{i=1}^\infty f_i(x), x\ne 0$. Does this implies that $F(0) = \sum_{i=1}^\infty f_i(0)$?

This is incorrect. Consider the functions $g_n$ given by
$$
g_n(x) = \left\{ \begin{matrix} 0 & x = 0 \\ nx & 0 \leq x \leq 1/n \\ -nx & -1/n \leq x \leq 0 \\ 1 & \text{otherwise} \end{matrix} \right.
$$
Visually, these functions look like this:

As $n$ grows, the slopes get steeper and the horizontal lines just get closer to the $y$-axis.
Let $f_1 = g_1$ and $f_{n+1} = g_{n+1} - g_n$. Since $g_{n+1} \geq g_n$, we know that the functions $f_n$ are nonnegative; since the $g_n$ are continuous, so are the $f_n$. Moreover, you can check that $\sum_{i=1}^n f_i(x) = g_n(x)$, so
$$\sum_{i=1}^\infty f_i(x) = \lim \limits_{i \rightarrow \infty} g_i(x).$$
Thinking about these functions graphically, it's pretty clear that $\lim \limits_{i \rightarrow \infty} g_i(x)$ is just the function which equals $1$ everywhere except at $x=0$, at which it is $0$. That is:
$$ \sum_{i=1}^\infty f_i(0) = 0, \qquad \sum_{i=1}^\infty f_i(x)=1 \text{ for } x \neq 0. $$
So if we take $F(x)=1$, the constant function, we have $\sum_{i=1}^\infty f_i(x) = F(x)$ for $x \neq 0$ but $\sum_{i=1}^\infty f_i(0) \neq F(0)$.
