measure zero point in differentiable “almost everywhere”

I am a new student in leaning real analysis and still confused about "almost everywhere.

Definition:

Points $x\in \mathcal{X}$ is true except for $x$ in some null set (with measure $0$) ", we way it is true a.e.

Example:

"Differentiable almost everywhere" means differentiable at every point outside a set of (Lebesgue) measure zero $\mu = 0$.

My question is:

Where is the set with measure zero in real line for example?

Hope to construct correct concept.

• What do you mean the measure at 0 is 0? $\mu$ is a set function, so you are measuring sets. It only makes sense to talk of $\mu(\{0\})$, which does make sense and is 0 if $\mu$ is Lebesgue. Secondly, what do you mean measure zero in the real line? Are you asking for a subset of $ℝ$ with measure 0? It makes no sense to talk of $\mu(\{±∞\})$ with $\mu$ a measure on $ℝ$, and you specifically mentioned the real line. – Calvin Khor Oct 7 '15 at 6:39
• I should ask where is the null set (measure 0) when discussing a function $f: \mathbb{R} \mapsto \mathbb{R}$. For example, $f(x) = x^2$ – sleeve chen Oct 7 '15 at 6:44
• Forget about measure zero for the moment, first do you understand what is a measure? – user99914 Oct 7 '15 at 6:46
• What prior knowledge do you have? From the question it sounds that you're out on deep water here. Do you know the Lebesgue Integral? Do you know what a measure is? If you don't you should probably start there. – skyking Oct 7 '15 at 6:53
• @sleevechen there are sets (some very weird) other than intervals with measure 0. I think it might be useful for you to work towards is the construction of Lebesgue measure via the infimum of open covers definition. – Calvin Khor Oct 7 '15 at 6:58

Differentiability a.e. means that there is some null set $N$ such that your function (say) $f$ is differentiable on $ℝ \setminus N$. This null set depends on $f$. One example is $$f(x) = |x|$$ This function has $N=\{0\}$. Another is the Cantor function, and it is differentiable a.e. with derivative $0$, and is not differentiable on the Cantor set. Yet another example is $$f(x) = 1$$ This function is differentiable a.e., in fact its differentiable everywhere but this doesn't stop you from setting $N=∅$.
• @sleevechen It means that every function that is "differentiable everywhere" is also "differentiable a.e.", going strictly by the definition. Why? Because we can find an $N$ that works in the definition of "differentiable a.e."; that $N$ is the empty set. – Calvin Khor Oct 7 '15 at 6:53