Measure of set where functions are simultaneously bounded by their respective averages This pertains to an exercise in Maggi's "Sets of Finite Perimeter and Geometric Variational Problems"-- specifically Exercise 1.14: 
If $u_k$, $1\leq k\leq N$ are summable on $\mathbb R^n$ with respect to a finite measure $\mu$, then $$ \mu\left(\bigcap_{k=1}^N\left\{ u_k\leqslant \frac{2N}{\mu(\mathbb R^n)}\int_{\mathbb R^n} u_k\mathrm{d}\mu\right\} \right)\geqslant \frac{\mu(\mathbb R^n)}{2} $$ 
The result seems intuitive enough; the set of points where the $u_k$ are collectively not too much bigger than their averages takes up ''most'' of space. However, I've had a bit of trouble trying to prove this in full generality, and have only been able to do it in the case of one $u_k$ which is positive. 
Indeed, let $A=\left\{ u\leqslant \frac{2}{\mu(\mathbb R^n)}\int_{\mathbb R^n} u\mathrm{d}\mu\right\}$. Then we compute that $$\int_{\mathbb R^n} u\geq \int_{\mathbb R^n\setminus A} u> \int_{\mathbb R^n\setminus A}\frac{2}{\mu(\mathbb R^n)}\int_{\mathbb R^n} u=2\left(1-\frac{\mu(A)}{\mu(\mathbb R^n)}\right)\int_{\mathbb R^n} u$$ 
If we assume to the contrary that $2\mu(A)<\mu(\mathbb R^n)$, then we reach the desired contradiction. 
Towards the general case (still considering only one function $u$), we can define the function $F(x):= \frac{2}{\mu(\mathbb R^n)}\int_{\mathbb R^n} u\mathrm{d}\mu-u(x)$, which is positive on $A$ (I wanted to reduce to the positive case because then I might be able to utilize monotonicity somewhere). I have been able to prove (under the assumption that $\mu(A)<\frac{\mu(\mathbb R^n)}{2}$) by integrating $f$ over $A$ that $$\int_A u < \int_{\mathbb R^n} u < \int_{\mathbb R^n\setminus A} u$$ Clearly, this also implies that the integral over $A$ is strictly negative, and that the integral over the complement is strictly positive. I am having a hard time seeing any potential insight from this.
Unfortunately, this is about all I've drummed up so far, and any helpful nudges in the right direction (especially towards the general case) would be very greatly appreciated. I suspect that a full proof is going to involve a function somehow cooked up from all of the $u_k$, and that it will be fundamentally different from the proof of the positive case (since monotonicity played the crucial role). Thanks!
Edit: I've now been able to prove the result for two positive, summable functions as follows: Set $A_k=\left\{ u_k\leqslant \frac{2N}{\mu(\mathbb R^n)}\int_{\mathbb R^n} u_k\right\}$ and $A=\cap A_k$. 
In general, for $u_1,\ldots, u_N$ positive and summable, we have for each $k$ that $$\int_{\mathbb R^n} u_k \geq \int_{\mathbb R^n\setminus A_k} u_k>2N\left(1-\frac{\mu(A_k)}{\mu(\mathbb R^n)}\right)\int_{\mathbb R^n} u_k.$$ Assuming without loss of generality that $u_k\neq 0$, we gather from this (by summing over the result for each $u_k$) that $$\sum_{k=1}^N \mu(A_k)>\left(N-\frac{1}{2}\right)\mu(\mathbb R^n).$$ Using now inclusion-exclusion in the case where $N=2$, we calculate that $$\mu(A_1\cap A_2)=\mu(A_1)+\mu(A_2)-\mu(A_1\cup A_2)>\left(2-\frac{1}{2}\right)\mu(\mathbb R^n)-\mu(\mathbb R^n)=\frac{1}{2}\mu(\mathbb R^n).$$
Unfortunately, the approach by inclusion-exclusion does not appear sharp enough to handle any other cases as far as I can tell, as I get trivial bounds from the approach.
 A: With some very helpful guidance from Dr. Maggi, the problem has been solved. Long story short, it can sometimes pay to ask what estimate you could use, and then see if you can prove it...
At any rate, we can prove the nice estimate that $\mu(A_k)\geqslant(1-\frac{1}{2N})\mu(\mathbb R^n)$. In retrospect, this should have been completely obvious since the larger $N$ becomes, the more room each $u_k$ can roam around and still be less than $2N$ times its average. 
If the estimate didn't hold, then we would find that $$\int_{\mathbb R^n} u_k\geqslant \int_{\mathbb R^n\setminus A_k} u_k>\frac{2N}{\mu(\mathbb R^n)}\mu(\mathbb R^n\setminus A_k)\int_{\mathbb R^n} u_k=2N\left(1-\frac{\mu(A_k)}{\mu(\mathbb R^n)}\right)\int_{\mathbb R^n} u_k $$ implying that $1>2N\left(1-(1-\frac{1}{2N}\right)=1$. Lo and behold, the estimate holds!
Now let $A=\cap A_k$. We compute that $$\mu(\mathbb R^n)-\mu(A)=\mu(\mathbb R^n\setminus A)=\mu\left(\bigcup( \mathbb R^n\setminus A_k)\right)\leqslant \sum\left( \mu(\mathbb R^n)-\mu(A_k)\right) $$ which yields $$\mu(\mathbb R^n)-\mu(A)\leqslant N\mu(\mathbb R^n)-N\left(1-\frac{1}{2N}\right)\mu(\mathbb R^n)=\frac{1}{2}\mu(\mathbb R^n) $$ as desired. 
Intuitively, that estimate would have been helpful because $$\mu(\mathbb R^n)-\mu(A)\leqslant \sum\left( \mu(\mathbb R^n)-\mu(A_k)\right) $$ gives $$\sum\mu(A_k)-(N-1)\mu(\mathbb R^n)\leqslant \mu(A), $$ and if the left hand side is at least as big as $\left(N\lambda-(N-1)\right)\mu(\mathbb R^n)\geqslant \frac{1}{2}\mu(\mathbb R^n)$, then we can take $\lambda\geqslant 1-\frac{1}{2N}$ and try to prove that $\mu(A_k)\geqslant \lambda\mu(\mathbb R^n)\geqslant\left(1-\frac{1}{2N}\right)\mu(\mathbb R^n)$. 
