Determine the set of real values where the given function $f$ does not equal the given function $g$. Then use Definition 1.2.3 to prove $f=g$ almost everywhere.
Definition 1.2.3: The functions $f$ and $g$ are said to be equal almost everywhere when $f(x)=g(x)$ for all $x$ in the functions domains, except possible for a set $x$ values that has measure zero.
The two functions are $f(x)=\begin{cases} 1 & x\in(0,1) \\ 0 & x\notin (0,1) \end{cases}$ and $g(x)=\begin{cases} 1 & x\in[0,1] \\ 0 & x\neq[0,1] \end{cases}$
I see that $f(1)\neq g(1) $ and $f(0)\neq g(0)$.
So is the set of real values where $f\neq g$ $\{0,1\}$?
And how to I show that $f=g$ almost everywhere using the definition?
This is what I have for determining measure zero: Let $\epsilon >0$ be given and let $S=\{0,1\}$. Then we can cover $S$ with the open intervals $I_1 = (-\frac{\epsilon}{4},\frac{\epsilon}{4})$ and $I_2 = (\frac{1-\epsilon}{4}, \frac{1+\epsilon}{4})$. $m(I_1)=0$ and $m(I_2)= \frac{\epsilon}{2}$.
Then $\sum_{n=1}^{2} m(I_n) = 0+\frac{\epsilon}{2} = \frac{\epsilon}{2} < \epsilon$
If your measure is Lebesgue measure $\lambda$, then $\lambda(\{0,1\})=0$ and the functions agree on the rest of the points, so you found your zero measure set. (I think you refer to Lebesgue measure as you are using open interval to compute the measure of your set, allegedly using the Lebesgue outer measure representation of measurable sets).
Of course on different measure spaces this might not be true. Let $(\mathbb R, \{\emptyset,\mathbb R,\{1\}, \mathbb R\backslash \{1\}\}, \delta_1)$ be your measure space, where $\delta_1(A)=1$ if $1\in A$ and $0$ otherwise. Then for every zero measure set $A$ in your $\sigma$-algebra (only $\mathbb R\backslash \{1\}$ and $\emptyset$), there exist a point $\mathbb R\backslash A$ where the two functions differ.