In this question, all measures are Lebesgue.
Let $ E $ be a measurable set. Is it true that for every $ \epsilon > 0 $, there is a step function $ \gamma : \mathbb{R} \longmapsto \mathbb{R} $ such that the set $ \left \{x \in \mathbb{R} | \gamma(x) \neq \chi_{E} (x) \right \} $ has measure less than $ \epsilon $? Here $ \chi_{E} : \mathbb{R} \longmapsto \mathbb{R} $ is the indicator function for $ E $, and so, it assumes values $ 1 $ when $ x \in E $ and $ 0 $ otherwise.
Note 1: I can do the proof when $ E $ has finite measure. The case $ m(E) = \infty $ is driving me nuts.
Note 2: I have another problem that may not be relevant here. Does a step function allow degenerate intervals? So, can a step function (which assumes finitely many values) be like $ f(x) = 2 $ for $ x < 2 \wedge x > 2$, and $ f(x) = 1 $ at $ x = 1 $?
Noting feels better than answering your own question, especially one as trivial as this.
So, let $ E = \bigcup _{ n \in \mathbb{Z} } [2n, 2n+1) $. If $ \gamma $ is any step function, then, it must be constant on an interval of the form $ (a, \infty) $ or $ [a, \infty) $. In that case, $ \gamma $ is not equal to $ \chi_{E} $ on either $ E \cap (a+1, \infty) $ or $ ( \mathbb{R} - E ) \cap ( a+ 1, \infty ) $, both of which have infinite measure.
Also, wikipedia mentions that sum and product of two step functions is a step function. Consider the functions $ s_{1} $ which assume value $ 1 $ for $ x < 0 $ and $ 2 $ for $ x \geq 0 $ and $ s_{2} $ which assumes value $ 3 $ for $ x \leq 0 $ and $ 1 $ on $ x > 0 $. The sum of these two functions is a step function which has a degenerate interval.