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I am stuck with the following problem. The problem is still unsolved.

Let $\left(X,\mu\right)$ be a measure space with a positive, finite measure $\mu$; and let $\left\{ f_{j}\in L^{\infty}\left(X\right)\right\}$ be a decreasing sequence converging pointwise to $f;$ $f_{j}\searrow f.$ Assume that $$ \intop_{X}f_{j}d\mu\geq-1. $$ Can we conclude that $$ \intop_{X}fd\mu\geq-1? $$

Thank you.

Hahn
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1 Answers1

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This is not true at all. For instance, consider $f_j(x)=-2X_{[0,j]}+2jX_{[j,j+1]}$. Then the integral of $f_j$ equals 0, whereas its pointwise limit is $f(x)=-2$, which is not even Lebesgue integrable. Of course, if you need $f$ to be integrable, you can still show that this still doesn't hold by taking $f_j(x)=-\Sigma_{k=0}^{j-1}2^{-j}+2X_{[j,j+1]}$, whose pointwise limit is $f(x)=-\Sigma_{k=0}^\infty 2^{1-j}X_{[j,j+1]}$, boasting an integral of -2.

Edit: I didn't notice the requirement that $f_j$ be decreasing. The given claim does hold - just apply the dominated convergence theorem for $\max{|f_1|,|f|}$ as the dominating function.

George K
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