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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.

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    You didn't need to write that $\int_X $ is just a Riemann integral. It can be proved that if a function is Riemann integrable then the Riemann and Lebesgue integral are the same.2012-04-05

1 Answers 1

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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 $\phi_{109}(x)=\color{red}{6}$.

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    See Edit. $ $ $ $2012-04-05