Show that $\lim_{j\to\infty}\int_{\mathbb{R}^n}f_jd\mathcal{L}^n=\int_{\mathbb{R}^n}f\mathcal{L}^n$ Let $g_j\le f_j\le h_j$ be three sequences of Lebesgue-integrable functions with $g_j\to g, f_j\to f, h_j\to h$ such that $g,h\in L^1(\mathbb{R}^n)$ and $\lim_{j\to\infty}\int_{\mathbb{R}^n}h_jd\mathcal{L}^n=\int_{\mathbb{R}^n}hd\mathcal{L}^n,\;\lim_{j\to\infty}\int_{\mathbb{R}^n}g_jd\mathcal{L}^n=\int_{\mathbb{R}^n}gd\mathcal{L}^n $.

Show that $f\in L^1(\mathbb{R}^n)$ and $\lim_{j\to\infty}\int_{\mathbb{R}^n}f_jd\mathcal{L}^n=\int_{\mathbb{R}^n}fd\mathcal{L}^n$.


My work so far: $f$ is integrable by the definition of the Lebesgue integral since both the negative (which I will denote by $f_{\_}$) and positive part of $f$ (which I will denote by $f_{+}$) are bounded by $g_{\_}$ and $h_{+}$.
My problem is the second part: I want to obtain something like $f_j\ge h_j+\epsilon$ and $f_j\le g_j-\delta$ for some $\epsilon,\delta&gt; 0$ in order to be able to use the fact that we can switch integral and limit for $g$ and $h$. I have already tried to make the convergence uniform by using Egorov locally and then gluing the result together but I don't know how to continue from here.
I don't think you should have uniform convergence. Here is my idea:
Let $f_{n-}$ denote the negative part of $f_n$ for every $n$. $f_{n-}$ is always a positive function. Thus, by Fatou's Lemma $$\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n\leq\liminf_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\limsup_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n.$$ The same follows analogously for $f_+$. The proof follows from the definition of the Lebesgue Integral.
I don't think you should have uniform convergence. Here is my idea:
Let $f_{n-}$ denote the negative part of $f_n$ for every $n$. $f_{n-}$ is always a positive function. Thus, by Fatou's Lemma $$\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n\leq\liminf_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\limsup_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n.$$ The same follows analogously for $f_+$. The proof follows from the definition of the Lebesgue Integral.