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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

$$ \limsup_{i \rightarrow \infty} \int_X \phi(z_i,x) dx < \infty $$


$$ \limsup_{i \rightarrow \infty} \phi(z_i,x) \text{ bounded almost everywhere on } X $$

Note: $\int_X (\cdot) dx$ is just a Riemann integral.

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From Fatou's Lemma, at least I have that $\liminf_{i \rightarrow \infty} \phi(z_i,x)$ is bounded almost everywhere. – Adam Apr 2 '12 at 18:25
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. – Beni Bogosel Apr 5 '12 at 6:28
up vote 4 down vote accepted

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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In your post, I guess bounded almost everywhere means finite almost everywhere. – Did Apr 4 '12 at 6:05
I meant that $\exists c$ such that $\limsup_{i \rightarrow \infty} \phi(z_i,x) \leq c \ \forall x \in \bar{X}$, where $\bar{X}$ is a set of measure $0$. – Adam Apr 4 '12 at 16:22
I'm not clear on the first part: $\phi_i(x) = n$ if $i = 2^n + k$ and $x \in [ \frac{k}{2^n}, \frac{k+1}{2^n} )$. Therefore it seems to me that $\limsup_{i \rightarrow \infty} \phi_i(x) $ is unbounded only for $x=0$ as the interval $[ \frac{k}{2^n}, \frac{k+1}{2^n} )$ goes to $0$ as $i$, and consequently $n$, goes to $\infty$. – Adam Apr 4 '12 at 17:39
Read again: when $i\to\infty$, for each $n$, $k$ goes from $0$ to $2^n-1$, then $n$ is replaced by $n+1$ and $k$ goes from $0$ to $2^{n+1}-1$, then $n+1$ is replaced by $n+2$, and so on. Hence for every $x$ in $[0,1)$ and every $n$, $\phi_i(x)\ne0$ for exactly one index $2^n\leqslant i\lt 2^{n+1}$, and for this index $i$, $\phi_i(x)=n$. The sequence of values of $\phi_i(x)$ looks like 1, 0, 2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0,... – Did Apr 4 '12 at 22:16
Sorry, I'm still not clear: too many things are "going to $\infty$". I would start saying that $n$ and $k$ are fixed numbers, so I don't understand why $n$ is replaced by $n+1$ and then $n+2$... In other words, it seems to me that, for fixed $x$, $\limsup_{i} \phi_i = 0$ because $i = 2^n + k$ cannot be satisfied when $i$ goes to $\infty$ – Adam Apr 4 '12 at 22:35

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