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I am asking on what seems to be a non-trivial variant of the popular question: If $f$ is Lebesgue integrable on $E$ and $\epsilon>0$, there exists $\delta>0$ such that for all measurable sets $X\subseteq E$ with $m(X)<\delta$, $|\int_X f\,d\mu|<\epsilon$.

I know how to solve the above question. What puzzles me is the below variant:

Q) Suppose $f$ is a measurable function on a measurable set $E\subset\mathbb{R}^n$ such that $\int_E |f|\,dx<\infty$. Prove that for any $\epsilon>0$, there exists a compact set $K\subseteq E$ such that $|\int_{E\setminus K} f(x)\,dx|<\epsilon$.

My main problem is how to come up with a compact $K$. I can come up with a closed $K$ by using Inner Approximation, so there exists closed $K$ such that $|E\setminus K|<\delta$.

However, since $E$ is not necessarily bounded, so I can't get that $K$ is compact.

Any hints? Thanks!


Update: I am thinking of using $E_M=E\cap B_M(0)$, and $E_M\nearrow E$, so by Monotone Convergence Theorem for Measure, there exists sufficiently large $M$ such that $|E|-|E_M|=|E\setminus E_M|<\epsilon/2$.

Take closed (and bounded thus compact) $K\subset E_M$ such that $|E_M\setminus K|<\epsilon/2$. Then $|E\setminus K|\leq|E\setminus E_M|+|E_M\setminus K|<\epsilon$.

Is that correct?

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  • $\begingroup$ Note that $f$ is `integrable'. So you don't have to worry about boundedness of $E$. $\endgroup$ – Will Kwon Oct 5 '16 at 14:50
  • $\begingroup$ Hmm.. Thanks. Any elaboration? I still don't get it.. $\endgroup$ – yoyostein Oct 5 '16 at 14:50
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    $\begingroup$ Cut-off and monotone convergence theorem will be helpful. $\endgroup$ – Will Kwon Oct 5 '16 at 14:51
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    $\begingroup$ Also since you are playing on Euclidean space, consider $K$ as an intersection with ball and $E$ $\endgroup$ – Will Kwon Oct 5 '16 at 14:53
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Let $E_k = E \cap [-k,k]^n$. The monotone convergence theorem implies that $$\int_{E_k} |f| \nearrow \int_E |f|$$ so that $$\int_{E \setminus E_k} |f| \to 0.$$

Now argue using a compact subset $K$ of $E_k$ rather than of $E$: $$\left| \int_{E \setminus K} f \right| \le \int_{E_k \setminus K}|f| + \int_{E \setminus E_k}|f|.$$

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