# Finite convergence of integral over a space implies convergence of integral over all measurable functions in space.

I'm trying to prove the following:

Suppose $\{f_n\}\in L^+$, $f_n \to f$ pointwise, and $\int f = \lim \int f_n < \infty$. Then $\int_Ef = \lim \int_E f_n$ for all $E$ in $\mathcal M$.

$L^+$ is the space of measurable functions from the space X to $[0, \infty]$. For this problem I am not able to assume dominated convergence or Egoroff's lemma, as those appear in the next section. The tools available are mainly results dealing with simple functions, the monotone convergence theorem, and Fatou's lemma.

My best guess is to define functions $g = fχ_E$ and $g_n = f_nχ_E$ (where $χ_E$ is the indicator function for $E$, i.e., 1 if x is in E and 0 otherwise) and try to make $∫g_n$ converge to $∫g$. The best I can come up with is that $\lim(g_n) = g$ pointwise, so by Fatou's lemma, $∫g$ $= \lim(g_n)$ $= \int\liminf(g_n)$ $≤ \liminf ∫g_n$, but this obviously isn't good enough, and I don't think there are any results that tell me about $\limsup\int g_n$, which seems like the most obvious way to make this inequality useful.

-
Is $L^+$ the space of positive integrable functions? If so use dominated convergence and the triangle inequality to prove $L^1$ convergence: $0 \leq |f_n - f| + f - f_n \leq 2f$. – Chris Janjigian Mar 1 '14 at 19:18
$$\underline\lim \int_E f_n \ge\int_E f =\int_X f -\int_{E^C} f\ge \int_X f -\underline\lim \int_{E^C} f_n=\overline\lim \Bigl(\,\int_X f_n -\int_{E^C}f_n\Bigr)=\overline\lim\int_E f_n$$ – David Mitra Mar 1 '14 at 19:37
$L^+$ is the space of all measurable functions from a given space X to [0, infinity]. I don't think I can use dominated convergence either; it's also in the next section. – Xindaris Mar 1 '14 at 22:59