If $U$ is a real measurable function s.t. $\int_{B_1(x)}\frac{1}{U(y)} dy\to 0$, does $\int_{B_1(x)}\frac{1}{U(y)+a} dy\to 0$, too? Let $U:\mathbb{R}^N\to\mathbb{R}$ a measurable function such that
$${\rm infess}_{\mathbb{R}^N} U(x)>-\infty.$$
Let $a\in\mathbb{R}$ be sufficiently large such that
$${\rm infess}_{\mathbb{R}^N} (U(x)+a)>0.$$
Finally, assume that
$$(I) \int_{B_1(x)}\frac{1}{U(y)} dy\to 0 \quad\mbox{ as } |x|\to +\infty.$$
My question is: assumption $(I)$ implies that also
$$\int_{B_1(x)}\frac{1}{U(y)+a} dy\to 0 \quad\mbox{ as } |x|\to +\infty?$$
If yes, how to prove it? I’ve been stuck for a while on it, I hope someone could help.
Thank you in advance!
 A: I assume infess$_{\mathbb{R}^N}$ means the 'essential infimum' taken over $\mathbb{R}^N$, and $B_1(x)$ means the open ball of radius $1$ centered at $x$. Please correct me if I misunderstood them.
In the simplest case where $N=1$ we already can construct a counterexample. The idea is that $U$ (hence $1/U$) may oscillate to change sign over the balls, causing the cancellation between the positive and negative integrals. For example, $\int_{(-1,1)}\sin(2\pi x)dx=0$. In fact I'm going to construct a counterexample based on the sine function.
Let $\phi:\left(0,2\pi\right)\to\mathbb{R}$ be the function defined by
\begin{align}
\phi(x):=\left\{
\begin{array}{ccl}
\frac{1}{2} & \text{if} & x\in\left(0,\frac{\pi}{6}\right)
\cup\left(\frac{5\pi}{6},\pi\right) \\
\sin x & \text{if} & x\in\left[\frac{\pi}{6},\frac{5\pi}{6}\right]
\cup\left[\frac{7\pi}{6},\frac{11\pi}{6}\right] \\
-\frac{1}{2} & \text{if} & x\in\left(\pi,\frac{7\pi}{6}\right)
\cup\left(\frac{11\pi}{6},2\pi\right) \\
\end{array}
\right.
\end{align}
Plotting the graph and you can see that I'm merely modifying the graph of sine function near the points where it is zero by some nonzero constants ($\pm 1/2$). Then, let $\Phi$ be the $2\pi$-extension of $\phi$ to the whole real line. (Actually I have not defined $\Phi$ on the set $Z:=\{x:sin x=0\}=\{k\pi:k\in\mathbb{Z}\}$, but this is a set of measure zero and you are free to define anything you like on it, since the integral won't be affected.) Finally, set
\begin{align}
U(x):=\frac{1}{\Phi(2\pi x)}
\end{align}
Then $U$ is a measurable function satisfying
\begin{align}
\text{infess}_{\mathbb{R}}U=-2>-\infty
\end{align}
(This is the reason why I modify the sine function to be nonzero near points of $Z$. Had this not been done and we naively set $\Phi(x)=\sin x$, $U$ would not be essentially bounded below.) Choosing, e.g. $a=3$ will guarantee that infess$_{\mathbb{R}}(U+a)>0$.
Now, the periodicity of $\Phi$, hence of $U$, will imply that condition $(I)$ holds, by forcing the stronger fact that
\begin{align}
\int_{B_1(x)}\frac{1}{U(y)}dy=\int^{x+1}_{x-1}\Phi(2\pi y)dy=0
\end{align}
for every $x\in\mathbb{R}$, since $\Phi(2\pi y)$ has period $1$, but the interval $B_1(x)=(x-1,x+1)$ has length $2$, and also due to the fact that
\begin{align}
\int^{\pi}_0\phi(x)dx=-\int^{2\pi}_{\pi}\phi(x)dx
\end{align}
However, on the other hand we can show that
\begin{align}
\frac{1}{U(y)+a}=\frac{\Phi(2\pi y)}{1+3\Phi(2\pi y)}
\end{align}
has a positive lower bound (says $c>0$) on $\mathbb{R}\setminus Z$, and so the integral
\begin{align}
\int_{B_1(x)}\frac{1}{U(y)+a}dy\geq c\cdot\text{length}B_1(x)=2c
\end{align}
is bounded below by a positive constant for all $x\in\mathbb{R}$, hence cannot tend to $0$ as $|x|\to+\infty$.
