Krdytkyu

I am faced with the following problem, I need to estimate the Lebesgue measure of the level set ${f>alpha}$ within a program. The function $f$ is approximated by a vector of arguments $(t_i)_{i=1}^n$ and a vector of corresponding values $(y_i)_{i=1}^n$. The only solution that comes to mind is to approximate the measure by $hat{lambda}=frac{1}{n}|{t_i:y_i>alpha}|$. This raised an interesting question for me. Given a measurable set $Asubset [0,1]$ and an equidistant partition $t_i=frac{i}{n}$ with $i=0,ldots,n$, I define the estimator $hatlambda_n(A):=frac{1}{n}|{t_i:t_iin A}|$. What conditions must I impose on $A$ such that
$$
lim_{ntoinfty}hatlambda_n(A)=lambda(A),, ?
$$

Also, if anyone has a suggestion on how to better approximate the measure of a level set given a discrete approximation, I would be very glad to hear from you. Thank you in advance

Define for every $E$ in the unit interval, $overline{mu}(E)= overline{lim}_{ntoinfty} hat{lambda}_n(E)$, where $hat{lambda}_n=frac{1}{n}sum_{i=1}^n delta_{t_i}$, as in your definition. Let $underline{mu}$ be the set function we get by replacing the limsup with liminf above. You may have noticed that $overline{mu}(I)=mathcal{L}^1(I)$ whenever $I$ is a subinterval (where $mathcal{L}^1$ is the Lebesgue measure).

Then given $E$ and a collection of intervals ${I_j}_j$ such that $Esubset bigcup_j I_j$, we have $$hat{lambda}_n(E)leq sum_j :hat{lambda}_n(I_j),$$ which gives after passing through limsup$$overline{mu}(E)leq sum_joverline{mu}(I_j)=sum_jmathcal{L}^1(I_j).$$ Taking the infimum over all such collection of intervals, we get begin{equation}overline{mu}(E)leq mathcal{L}^1(E),;;;;;;;quadquadquad(1)end{equation} for any subset of the unit interval.

Let $Ksubset [0,1]$ be compact. Given $ninmathbb{N}$, Let $Usubset [0,1]$ be any open set containing $K$. Let $U_n$ be the open set which is the union of the minimal collection of intervals of the form $(t_i-frac{1}{n^2},t_i+frac{1}{n}+frac{1}{n^2})$ needed to cover $U$. Then $$mathcal{L}^1(K)leq mathcal{L}^1(U)leq sum_{t_iin U_n}(1/n + 2/n^2)=(1+1/n^2)frac{1}{n}|{t_i:t_iin U_n}|=(1+1/n^2)hat{lambda}_n(U_n)leq (1+1/n^2)mathcal{L}^1(U_n). $$ Taking liminf over $n$ and a decreasing collection of $Usupset K$, due to sandwiching and (1) we get, $$mathcal{L}^1(K)leq underline{mu}(K).quadquadquadquadquad (2)$$

If $E$ is an $F_{sigma}$ set, then it is the union of an increasing sequence of compacts ${K_i}_i$. We have by (1) and (2),$$overline{mu}(E)leq mathcal{L}^1(E)=sup_i mathcal{L}^1(K_i) leq sup_iunderline{mu}(K_i)leq underline{mu}(E).$$

So, a sufficient condition for the equality to hold is that the set be $F_{sigma}$.

Note that the outer measure $overline{mu}$ is not Borel regular, because if it were, it would have to be (by the uniqueness of Haar measure on $[0,1)$) the Lebesgue measure, which it is not because the measure of irrationals by $overline{mu}$ is zero. However, you do get the Lebesgue measure as the weak limit of a subsequence of the sequence $hat{lambda}_n$.

Regarding a better way to approximate, I don't really have anything meaningful to say (supposing what I wrote above in some way was). If the function has some regularity, there may be better ways of approximating the size of the level sets than taking a rough average, for example, taking a discrete mollification.