Sequence of non-integrable functions converging to an integrable function Is it possible to have the following:

A sequence $(h_n) \rightarrow h$ uniformly where each $h_n$ is not integrable but $h$ is integrable.

 A: If you mean Riemann integrable on a bounded interval, then consider
$$h_n(x) = \begin{cases}\frac1{n}, \,\, x \in \mathbb{Q} \cap [0,1]\\ 0, \, \, \,\,x \notin \mathbb{Q} \cap [0,1]  \end{cases}$$
A: Take $h_n =1/n$ on the real line. Then $h_n \rightarrow 0$ uniformly, and $0 \in L^1$.
A: Example 1.There is a standard textbook example of a non-measurable $S_q \subset (0,1]$ such that $\forall q\in Q\cap (0,1]$ the set $(0,1]\cap S$ is non-measurable. For $n\in N$ let $f_n (x)=0$ for $x\not \in (0,1/n),$ and $f(x)=0$ for $x\not \in S,$ and $f(x)=1/n$ for $x\in (0,1/n)\cap S.$ Then $f_n$ is not Lebesgue-integrable because $f_n$ is not Lebesgue-meaurable, but $f_n\to 0$ uniformly.
Example 2. Let  be any non-measurable set. Let $1_S$ be the characteristic function of $S.$ Let $f_n=(1/n)1_S$ for $n\in N.$ 
A: Example 1.There is a standard textbook example of a non-measurable $S_q \subset (0,1]$ such that $\forall q\in Q\cap (0,1]$ the set $(0,1]\cap S$ is non-measurable. For $n\in N$ let $f_n (x)=0$ for $x\not \in (0,1/n),$ and $f(x)=0$ for $x\not \in S,$ and $f(x)=1/n$ for $x\in (0,1/n)\cap S.$ Then $f_n$ is not Lebesgue-integrable because $f_n$ is not Lebesgue-meaurable, but $f_n\to 0$ uniformly.
Example 2. Let  be any non-measurable set. Let $1_S$ be the characteristic function of $S.$ Let $f_n=(1/n)1_S$ for $n\in N.$ 
Example 3. Let $f_n(x)=1/n$ for all $x\in R.$ Then $\int_Rf_n(x)\;dx=\infty$ for each $n$ but $f_n\to 0$ uniformly.
Counterpoint: Let $T$ be a  measurable set with measure $m(T)<\infty.$ Let $f_n:T\to R$ be a measurable function for each $n\in N.$ If $(f_n)_n$ converges uniformly on $T$ to an integrable $f:T\to R$ then for all but finitely many $n$ we have $-m(T)\leq \int_T (f_n(x)-f(x))\;dx\leq m(T),$ which implies $$-m(T)+\int_Tf(x)\;dx\leq \int_Tf_n(x)\;dx\leq m(T)+\int_Tf(x)\;dx$$ for all but finitely many $n$. So $f_n$ is integrable for all but finitely many $n$. (If you like you can extend $f$ and each $f_n$ by letting $f(x)=f_n(x)=0$ for $x\not \in T.$)
