0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Calvin Khor Oct 7 '15 at 6:39
  • $\begingroup$ I should ask where is the null set (measure 0) when discussing a function $f: \mathbb{R} \mapsto \mathbb{R}$. For example, $f(x) = x^2$ $\endgroup$ – sleeve chen Oct 7 '15 at 6:44
  • $\begingroup$ Forget about measure zero for the moment, first do you understand what is a measure? $\endgroup$ – user99914 Oct 7 '15 at 6:46
  • 1
    $\begingroup$ 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. $\endgroup$ – skyking Oct 7 '15 at 6:53
  • 1
    $\begingroup$ @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. $\endgroup$ – Calvin Khor Oct 7 '15 at 6:58
2
$\begingroup$

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 $N=∅$.

$\endgroup$
  • $\begingroup$ What does the last sentence mean? $\endgroup$ – sleeve chen Oct 7 '15 at 6:52
  • 1
    $\begingroup$ @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. $\endgroup$ – Calvin Khor Oct 7 '15 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.