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.

  • 1
    $\begingroup$ Do you assume that $f_j \in L_1$ ... or would you allow $\int f_j\;d\mu = +\infty$ for all $j$ ? In that case you could have $\int f\;d\mu < -1$. $\endgroup$ – GEdgar May 12 '18 at 14:53
  • $\begingroup$ Hi, I assume the measure is finite, which I forgot to add in. Since $\mu(X)<\infty$, it implies that each $f_{j}\in L^{1}$. Thank. $\endgroup$ – Hahn May 12 '18 at 17:44
  • $\begingroup$ Now also asked on MO: Monotone convergence of a integral under a decreasing sequence. $\endgroup$ – Martin Sleziak May 13 '18 at 8:23
  • $\begingroup$ Finiteness of the measure is an important assumption. Without that you can give an example where $\int f$ is not even defined. $\endgroup$ – Kavi Rama Murthy May 13 '18 at 9:18

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.

  • $\begingroup$ It seems that the sequence in your example is not decreasing. $\endgroup$ – Hahn May 12 '18 at 13:29
  • $\begingroup$ Oh, that is accurate. Do excuse my lack of attention. $\endgroup$ – George K May 12 '18 at 13:39

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