There is a subtle difference in defining Lebesgue integrals in Real analysis textbooks:
I) The approach of Royden & Fitzpatrick (in “Real analysis” 4th ed), Stein & Shakarchi (in “Real Analysis: Measure Theory, Integration, And Hilbert Spaces”)
Firstly, it defines Lebesgue integrability and Lebesgue integral for a bounded function (not necessarily measurable) on a domain of finite measure. A bounded function needs to be Lebesgue integrable first (the upper and the lower Lebesgue integral agree), then the integral can be defined to be this common value. The authors’ motivation is try to define “Lebesgue integrability” like “Rieman integrability”: upper integral equals lower integral.
However, unfortunately, the upper and lower Lebesgue integrals don’t agree for an arbitrary Lebesgue integrable function, so when the authors move to functions in general (not necessarily bounded), they still have to go back to the requirement "measurable". This sudden appearance of "measurability" is not natural.
(Note that the upper/lower Darboux sum in the definition of Rieman integrability can be viewed as step functions, which are a special case of simple functions. So “upper/lower Rieman (Darboux) integral” is a special case of “upper/lower Lebesgue integral”)
II) The approach of Folland (in “Real Analysis: Modern Techniques and Their Applications”), Bruckner & Thomsom (in “Real analysis”), Carothers (in “Real analysis”), etc.
The construction requires a function to be measurable, and defines the Lebesgue integral to be the upper Lebesgue integral, and when the integral is finite the function is said to be Lebesgue integrable.
This approach doesn’t immediately show how Lebesgue integral convers Rieman integral, so later on, the author proves that in the case a function is bounded and the domain of integration is of finite measure: the upper Lebesgue integral equals to the lower Lebesgue integral, which means Lebesgue integral is reduced to Rieman integral.
Not necessarily.
Consider a Smith-Volterra-Cantor set, a compact and nowhere dense set $C \subset [0,1]$ with positive Lebesgue measure. Let $C_n$ be the finite union of closed intervals remaining at stage $n$ of the construction, so that $C_1 \supset C_2 \supset \dots$ and $C = \bigcap_n C_n$.
Set $f_n = 1_{C_n}$ and $f = 1_C$ to the corresponding indicator functions, so we have $f_n \to f$ pointwise everywhere. Now I claim that for each $x \in C$, we do not have $f_n \to f$ locally uniformly at $x$. Let $U$ be any neighborhood of $x$ and let $n$ be arbitrary. Since $x \in C_n$, which is a union of closed intervals, we have $x \in I \subset C_n$ for some closed interval $I$. But $C$ is nowhere dense, so there exists some $y \in (I \setminus C) \cap U$. Hence we have $y \in U$, $f_n(y) = 1$, and $f(y) = 0$, so $\sup_{z \in U} |f_n(z) - f(z)| = 1$, and so the sequence $f_n$ does not converge to $f$ uniformly on $U$. This holds for all $x \in C$, and $C$ has positive measure.
Best Answer
Maybe you mean something like this?: let $f_n(x):=\chi_{[n,n+1]}(x)$ for $n\in\Bbb N$. Then clearly $f_n\in\mathcal L_1(\Bbb R)$ and $(f_n)\to 0$ point-wise, but
$$\lim_{n\to\infty}\int_{\Bbb R} f_n(x)\,\lambda(dx)=1\neq\int_{\Bbb R} \lim_{n\to\infty}f_n(x)\,\lambda(dx)=0$$