Bounded $\limsup$ integral implies $\limsup$ bounded almost everywhere? Consider $z \in \mathbb{R}^n$ and $\{ z_i \}_{i=1}^{\infty}$ with $z_i \rightarrow z$.
Let $\phi: \mathbb{R}^n \times X \rightarrow \mathbb{R}_{\geq 0}$. $X$ is unbounded.
I'm wondering if
$$ \limsup_{i \rightarrow \infty} \int_X \phi(z_i,x) dx < \infty $$
implies
$$ \limsup_{i \rightarrow \infty} \phi(z_i,x) \text{ bounded almost everywhere on } X $$
Note: $\int_X (\cdot) dx$ is just a Riemann integral.
 A: No, this does not imply that. For a counterexample, let $X=[0,1]$ and define $\phi_i(x)=n$ if $i=2^n+k$ with $0\leqslant k\lt2^n$ and $k\leqslant 2^nx\lt k+1$, and $\phi_i(x)=0$ otherwise. 
Then $\limsup\limits_{i\to\infty}\phi_i(x)=+\infty$ for every $x$ in $[0,1)$. If $i=2^n+k$ with $0\leqslant k\lt2^n$, then $\int\limits_X\phi_i=n/2^n$, hence $\lim\limits_{i\to\infty}\int\limits_X\phi_i=0$.
Edit Let us compute $\phi_{35}(x)$ for $x=1/\sqrt2$. Note that $35=2^\color{red}{5}+\color{blue}{3}$ and $2^\color{red}{5}x\approx\color{green}{22}.6$. Since $2^\color{red}{5}x$ is not in $[\color{blue}{3},\color{blue}{3}+1)$, $\phi_{35}(x)=0$. 
Furthermore, $\color{green}{22}\leqslant2^\color{red}{5}x\lt\color{green}{22}+1$ and $2^\color{red}{5}+\color{green}{22}=\color{purple}{54}$ hence the only nonzero $\phi_i(x)$ for $2^\color{red}{5}=32\leqslant i\leqslant63=2^{\color{red}{5}+1}-1$ is $\phi_{\color{purple}{54}}(x)=\color{red}{5}$. 
The sequence $(\phi_i(x))_{1\leqslant i\leqslant35}$ is
$$
0|0,\color{red}{1}|0,0,\color{red}{2},0|0,0,0,0,0,\color{red}{3},0,0|0,0,0,0,0,0,0,0,0,0,0,\color{red}{4},0,0,0,0|0,0,0,0
$$
and the two next nonzero values of $\phi_i(x)$ are $\phi_{54}(x)=\color{red}{5}$ and $\phi_{109}(x)=\color{red}{6}$.
