Problem about limit of Lebesgue integral over a measurable set

Question

This is actually problem 4T of Bartle's book "The elements of integration and Lebesgue measure".

Let $f_n$, $f$ be nonnegative measurable functions on $\mathbb{R}$ such that $f_n\to\ f$ for every real number (pointwise convergence). Suppose that $lim_{n\to \infty}\int_\mathbb{R}f_n=\int_\mathbb{R}f$.

Show that if $\int_\mathbb{R}f<\infty$, then $lim_{n\to \infty}\int_Ef_n=\int_Ef$ for every measurable set $E\subset\mathbb{R}$

Important theorems that I'm allowed to use: Theorem of Monotone Convergence, Fatou's Lemma, and Theorem of Dominated Convergence. I've been attempting to use dominated convergence by trying to define a sequence of functions bounded by f on $E$, but I haven't been able to come up with such a sequence.

Any hints would be greatly appreciated.

Answer 1

For any measurable set $E \subset \mathbb{R},$ we have $\displaystyle \int_Ef \leqslant \int_{\mathbb{R}}f < \infty$ and $\displaystyle \int_{E}f_n \leqslant \int_{\mathbb{R}}f_n \to \int_{\mathbb{R}}f$.

Hence, for sufficiently large $n$, we have

$$\displaystyle \int_Ef_n < \infty.$$

Using Fatou's Lemma,

$$\int_E f = \int_E \lim f_n=\int_E \liminf f_n \leqslant \liminf \int_E f_n$$

and reverse Fatou's Lemma,

$$\liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant \int_E \limsup f_n = \int_E \lim f_n = \int_Ef.$$

Hence,

$$\tag{*}\int_Ef \leqslant \liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant\int_Ef,$$

and

$$\liminf \int_E f_n = \limsup\int_E f_n=\lim \int_E f_n = \int_Ef.$$

Update:

We can avoid the argument based on reverse Fatou as follows.

Note that since $f_n \to f$ and $\int_{\mathbb{R}} f_n \to \int_{\mathbb{R}} f$ we have

$$\begin{align}\limsup \int_E f_n &= -\liminf \left(-\int_E f_n \right) \\&= -\liminf \left(\int_{\mathbb{R}\setminus E} f_n-\int_{\mathbb{R}} f_n \right) \\ &= -\liminf \int _{\mathbb{R}\setminus E} f_n+\liminf \int_{\mathbb{R}} f_n \\ &\leqslant -\int_{\mathbb{R} \setminus E} \liminf f_n + \int_{\mathbb{R}} f \\ &= -\int_{\mathbb{R} \setminus E} f + \int_{\mathbb{R}} f \\ &= \int_E f\end{align} $$

The chain of inequalities (*) now follows.