Here is a more readable version of the question:

Prove that for all functions $ g: mathbb {N} to mathbb {R} ^ { geq 0} $and all the numbers $ a in mathbb {R} ^ { geq 0} $, if $ g in Omega (1) $ then $ a + g in Theta (g) $

What I've done so far:

In order to $ a + g in Theta (g) $. $ a + g in mathcal {O} (g) Keil a + g in Omega (g) $,

If we expand our assumption, we get:

$ exists c_1, n_1 in mathbb {R} ^ {+}, forall n in mathbb {N}, n geq n_1 implies g (n) geq c_1 $

**To prove $ a + g in Omega (g) $**::

If we expand the definition, we get:

$ exists c_2, n_2 in mathbb {R} ^ {+}, forall n in mathbb {N}, n geq n_2 implies a + g (n) geq c_2 * g (n) $

To let $ c_1 = 1 $ and $ n_2 = n_1 $, To let $ n in mathbb {N} $, Accept $ n geq n_2 $, To prove $ a + g (n) geq c_2 * g (n) $,

$ g (n) = g (n) \ implies g (n) geq g (n) \ Leftrightarrow c_1 * g (n) geq c_1 * g (n) \ Leftrightarrow g (n) geq g (n) text {(since $ c_1 = 1 $)} \ Left right arrow a + g (n) geq g (n) text {(enlargement of the left side since $ a in mathbb {R} ^ { geq 0} $)} $

**To prove $ a + g in mathcal {O} (g) $**::

If we expand the definition, we get:

$ exists c_3, n_3 in mathbb {R} ^ +, forall n in mathbb {N}, n geq n_3 implies a + g (n) leq c_3 * g (n) $

I'm fighting here because I'm not really sure what its value is $ c_3 $ I should use or how to derive one. (I tried to use my assumption of $ g (n) geq c_1 $ but I don't really know where to go from there). Any help is greatly appreciated and I apologize for any formatting errors in advance. Thank you very much.