In asymptotic notation how to prove that $\mathcal{O}(g(n))\subseteq\mathcal{O}(f(n))\implies\mathcal{O}(f(n)+g(n))=\mathcal{O}(f(n))$ I have to prove that
 $$\mathcal{O}(g(n))\subseteq\mathcal{O}(f(n))\implies\mathcal{O}(f(n)+g(n))=\mathcal{O}(f(n))$$ 
The functions are non-negative.

Clarification: 
$$\mathcal{O}(f(n)+g(n))=\mathcal{O}(f(n))$$ 
it means 
$$\mathcal{O}(f(n)+g(n))\subseteq\mathcal{O}(f(n))\land\mathcal{O}(f(n))\subseteq\mathcal{O}(f(n)+g(n))$$

I try to explain my argument, i want to get on one side $f(n)+g(n)\in\mathcal{O}(f(n))$ and on the other: $f(n)\in\mathcal{O}(f(n)+g(n))$ in order to conclude that $\mathcal{O}(f(n)+g(n))=\mathcal{O}(f(n))$ true.

I start with the hypothesis:

$\mathcal{O}(g(n))\subseteq\mathcal{O}(f(n))\iff g(n)\in\mathcal{O}(f(n))$
$\iff(\exists c_1\in\mathbb{R}_{&gt;0},n_0\in\mathbb{N}:g(n)\leq c_1f(n),\forall n\geq n_0) \implies$ 
$\{$adding $f(n)$ to both members of the inequality$\}$ $(\exists c_1\in\mathbb{R}_{&gt;0},n_0\in\mathbb{N}:g(n)+f(n)\leq c_1f(n)+f(n),\forall n\geq n_0)\implies\quad(\exists c_1\in\mathbb{R}_{&gt;0},n_0\in\mathbb{N}:g(n)+f(n)\leq (c_1+1)f(n),\forall n\geq n_0)\implies \{\textrm{taking}\quad c_1^{'}=(c_1+1)\}\quad  (\exists c_1^{'}\in\mathbb{R}_{&gt;0},n_0\in\mathbb{N}:g(n)+f(n)\leq c_1^{'}f(n),\forall n\geq n_0)\iff \{by\quad definition\quad of \quad Oh-big \quad order\}\quad f(n)+g(n)\in\mathcal{O}(f(n))$

Now to achieve $f(n)\in\mathcal{O}(f(n)+g(n))$ start from the trivial proposition: $\mathcal{O}(f(n))\subseteq\mathcal{O}(f(n))$:

$\mathcal{O}(f(n))\subseteq\mathcal{O}(f(n))\iff f(n)\in\mathcal{O}(f(n))\iff(\exists c_2\in\mathbb{R}_{&gt;0}, n_0\in\mathbb{N}:f(n)\leq c_2f(n),\forall n\geq n_0)\implies$ $\{$adding $g(n)$ to right member of inequality$\}\quad (\exists c_2\in\mathbb{R}_{&gt;0}, n_0\in\mathbb{N}:f(n)\leq c_2f(n)+g(n),\forall n\geq n_0)$

At this point I'm stuck I don't know how to continue in order to achieve $f(n)\in\mathcal{O}(f(n)+g(n))$ to finish then the proof...

Thanks to everyone who read this :)
I think you are pretty much done.

$$
g(n) \ge 0 \implies f(n) \le f(n) + g(n), \quad \forall n \in \mathbb{N}
$$

So pick $c = 1 \in \mathbb{R}^+$ and $n_0 = 1 \in \mathbb{N}$, then
$$
f(n) \le c\cdot(f(n) + g(n)), \quad \forall n\ge n_0
$$