Property of Lebesgue measurable functions 
Prove that two continuous functions $f(x)$ and $g(x)$ are equivalent
  in terms of Lebesgue measure only when $f(x) \equiv g(x)$.

I'm afraid I'm not sure what is being asked here. In case this question is actually normal could you please explain?
 A: In Lebesgue theory, functions are equivalent if they are equal almost everywhere with respect to the Lebesgue measure, that is if they are equal up to a set of Lebesgue measure zero. For instance,
$$f(x)=\chi_Q(x)$$ is, in Lebesgue sense, equal to
$$g(x)= 0\quad \forall x.$$ Here, we denote by $\chi$ the characteristic function and by $Q$ the set of rational numbers (and remember that $Q$ is of Lebesuge measure zero).
In your case we have continuous functions and now the claim is that if they are identical with respect to Lebesgue measure (that is almost everywhere), that then they are identical for every $x$.
To prove this you can contradict the statement.
A sketch of proof:
Let $N = \{x : f(x) \neq g(x)\}$, then $N$ is of measure zero. Suppose for the sake of argument that $N$ is non-empty and we can pick $x \in N$. Now choose $\{x_k\} \notin N$ such that $x_k \to x$, then $$f(x) = f(\displaystyle\lim_{k \to \infty} x_k) = \displaystyle\lim_{k \to \infty} f(x_k) = \displaystyle\lim_{k \to \infty} g(x_k) = g( \displaystyle\lim_{k \to \infty} x_k) = g(x)$$ Therefore $x \notin N$. Hence we conclude that $N = \emptyset$. 
A: The set $\{x\mid f(x)\ne g(x)\}$ is open because $f,g$ are continuous. If $f\ne g$, then is is additionally not empty. Every nonempty open set has positive measure, hence we cannot have $f(x)=g(x)$ a.e. unless $f\equiv g$.
