# Limit of a measurable function and the Lebesgue integral

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 $n$?

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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 $N \rightarrow \infty$.

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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? –  Domonic Mei Nov 5 '12 at 15:53
Actually, if $|g| < \infty$ a.e., then $|E_N^c| \rightarrow 0$ as $N \rightarrow \infty$. –  Kelson Vieira Nov 5 '12 at 16:13