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I have problem with the exercise that follows.

Let $(z_m)_m \in R^n$ so that $\Vert z_m \Vert \rightarrow \infty$ when $m\to \infty$. Let $f:R^n \rightarrow [-\infty;+\infty]$ integrable.
Show that if $K \subset R^n$ is a compact $\lim_{m\rightarrow\infty} \int_{z_m+K}f d\lambda=0$.

I manage to find the result for $\vert f \vert$. But I can find a way to get to result for f, as asked in the exercise.I thought about using the result for $\vert f \vert$ but then I stuck..so maybe there's another way If someone can help me.
Update:
Also because then I've another problem related to the first one and I found a way to show it related to the exercise before with $\vert f \vert$ instead of $f$.But maybe there's a better way to show it.

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If you have the resul for $|f|$, use the fact: $$0\leq\left|\int_E f\right|\leq \int_E |f|$$ –  leo Nov 4 '12 at 1:54

2 Answers 2

up vote 0 down vote accepted

Here's a contradictory approach:

Suppose $\lim_{m\rightarrow\infty} \int_{z_m+K}|f| d\lambda \neq 0$. Then there exists $\delta>0$ such that $\int_{z_m+K}|f| d\lambda \geq \delta$ on some subsequence $z_{m_k}$. Since $\|z_{m_k}\| \to \infty$, and $K$ is bounded, we can choose a further subsequence so that the sets $z_{m_{k_i}}+K$ are disjoint. Then letting $A_n = \cup_{i=1}^n (z_{m_{k_i}}+K) $, we have $\int |f|d \lambda \geq \int_{A_n} |f|d\lambda \geq n \delta$, which is a contradiction since $f$ was assumed integrable.

Hence we have $\lim_{m\rightarrow\infty} \int_{z_m+K}|f| d\lambda = 0$ (marginally stronger than required).

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I'm sorry it was not clear that I got the results for $\vert f \vert$ but I can't get the same for f. –  Laura Nov 4 '12 at 0:32
1  
Well, since $|\int f| \leq \int |f|$, the result follows. Another way would be to let $f_+(x) = \max(f(x),0)$ and similarly $f_-(x)=-\min(f(x),0)$. Then $f = f_+-f_-$, $|f| = f_+ + f_-$, from which it follows that both $f_+,f_-$ are integrable & non-negative. Then prove the result separately for $f_+$ and $f_-$ (note $|f_+| = f_+, |f_-| = f_-$, so you are able to do that) and combine them together. –  copper.hat Nov 4 '12 at 2:44

Let $f_n(x):=|f(x)|\chi_{|x|\leq n}$. This forms a increasing sequence of integrable functions. By the monotone convergence theorem, $\lVert f-f_n\rVert_1\to 0$. This gives $$\int_{z_m+K}|f(x)|dx\leq \int_{z_m+K}|f(x)-f_n(x)|dx+\int_{z_m+K}|f_n(x)|\\\leq\lVert f-f_n\rVert_1+\int_{(z_m+K)}|f(x)|\chi_{B(0,n)}(x)dx.$$ For a fixed $n$, as $|z_m|\to +\infty$, $B(0,m)$ and $z_m+K$ are disjoint for $m\geq N_n$. This implies $$\limsup_{m\to +\infty}\int_{z_m+K}|f(x)|dx\leq \lVert f-f_n\rVert_1.$$ As $n$ is arbitrary, the result follows.

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