Suppose $\{f_n\}$ is a sequence of lebesgue measurable functions such that $f_n\rightarrow f$, except on a set of measure $0$, as $n\rightarrow\infty$, and $|f_n(x)|\leq g(x)$, where $g$ is integrable.

Denote $E_N = \{x: |x| \leq N, g(x)\leq N\}$. If I can prove that $m(E_N^c)\rightarrow\infty$ as $n\rightarrow\infty$, does that tell me anything about the lebesgue integral on that set?

Specifically, can I determine that $$\int_{E_N^c} |f_n -f| \leq \epsilon$$

for some $\epsilon > 0$, and all large $n$?


First, if $g\geq 0$ is integrable, then given $\varepsilon>0$, exist $N$ such that

$$\int_{E_N^c} g < \varepsilon.$$

Indeed, define $g_k:=g\chi_{E_k}$. Thus, $g_k\nearrow g$ and, by monotone convergence theorem, exist $N$, such that, $$0\leq \int (g - g_N)< \varepsilon.$$ Since $1-\chi_{E_N} = \chi_{E_N^c}$, we obtain the estimate.

Finally, note that $|f|\leq g$ and therefore $|f_n - f|\leq 2g$. Then, we conclude that exist $N$, such that $$\int_{E_N^c} |f_n - f|\leq 2\int_{E_N^c} g < 2\varepsilon,\ \ \ \ \forall n$$

Notice that $E_N \nearrow \mathbb{R}^d$, and so $|E_N^c|\rightarrow 0$, as $N \rightarrow \infty$.

  • $\begingroup$ Ahh, I think I'm wrapping my head around it now. So is the fact that $E_N^c\rightarrow 0$ necessary for the approximation of $g$, or literally just an upshot of how I defined it? $\endgroup$ – madisonfly Nov 5 '12 at 15:53
  • 1
    $\begingroup$ Actually, if $|g| < \infty$ a.e., then $|E_N^c| \rightarrow 0$ as $N \rightarrow \infty$. $\endgroup$ – Kelson Vieira Nov 5 '12 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.