Inequality for integral => Inequality for integrand I have that for any measurable set $\Omega\subset\mathbb{R}^d$ with $|\Omega|<\infty$ 
\begin{align}\sqrt{\int_\Omega f(x) dx }\leq \sqrt{c\cdot|\Omega|} + \sqrt{\int_\Omega g(x) dx }.\end{align}
How can I conclude that $\sqrt{f(x)} \leq \sqrt{c} + \sqrt{g(x)}$ for almost every $x\in\mathbb{R}^d$.
Cheers, Peter
 A: If $f$ and $g$ are continuous, then easily we can get the result for all $x \in R^d$.
Then we consider the case where $f$ and $g$ are only measurable. Let A be a measurable subset of $R^d$ with $|A| < +\infty$, then by Luzin's theorem, for any $\epsilon > 0$, there exists a closed set $E \subset A$ with $|A -E| < \epsilon$, such that $f$ and $g$ restricted on $E$ is continuous. 
Lemma((Proof at the end)) :We can prove that for almost every $x \in E$, we have $B(x, r) \cap E$ is of positive measure for any $r > 0$.(when $d=1$, those points for which this is not true are at most countable)
Then we can easily show that for most every point in $E$, the result is true. Thus the points in A for which the result is not true is of measure less than $\epsilon$. Since $\epsilon$ is arbitrary, we get that the result it true for almost every point in $A$.
Then use the fact that $R^d$ is countable union of sets like $A$ to conclude. 
Proof for the lemma:
Denote $U = \{x \in E, s.t. |B(x,r_x)\cap E| = 0 \text{ for some } r_x\}$. Obviously, we have $U \subset \cup_{x \in U}B(x, r_x)$.
For $q \in (\cup_{x \in U}B(x, r_x))\cap Q^d $, define $l_q$ such that $B(q, l_q) \subset B(x, r_x)$ for some $x \in U$. Then $U \subset \cup_{q \in B(x, r_x))\cap Q^d}B(q,l_q)$.
If $U$ is of positive measure, since $\{q, q \in B(x, r_x))\cap Q^d\}$ is countable, there exists some $q_0$ such that $U \cap B(q_0, l_{q_0})$ is of positive measure. Then there exists some $x \in U$ such that $U\cap B(x,r_x)$ is of positive measure. Since $U$ is part of $E$, it is contradictory with the definition of $B(x,r_x)$
