# Problem about limit of Lebesgue integral over a measurable set

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.

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.

• @salvador: You're welcome. – RRL Mar 14 '15 at 5:48
• I think this step $$\limsup\int_E f_n \leqslant \int_E \limsup f_n$$ is incorrect ..... or in case I am wrong just tell me why is this inequality correct? – Idonotknow Oct 15 '19 at 2:15
• @Idonotknow: The reverse Fatou lemma involving $\limsup$ holds if there is a dominating integrable function $g$ such that $f_n \leqslant g$ for all $n$. I think this can be found given the other hypotheses. As this is over 4 years old I have to spend some time trying to reconstruct my thinking that led me to believe it to be true. – RRL Oct 15 '19 at 6:20
• So, I am sorry for this question I know the question is long ago but I can not see anything about reverse Fatou 's Lemma in Royden 4th edition "Real Analysis" – Idonotknow Oct 15 '19 at 7:11
• Update added. I still think I had found the necessary dominating function to apply the reverse Fatou lemma, but I have to work on that further. – RRL Oct 15 '19 at 7:47