Integral with infinite limits as limit of integrals on sets of finite measure If $f(x,y) \rightarrow 0$ as $y \rightarrow 0$, then it can be shown that $|\int_{0}^{M} f(x,y)dx| \rightarrow 0 $ as $y \rightarrow 0$ for all finite $M \geq 0$. Could I also conclude that $|\int_{0}^{\infty} f(x,y)dx| = \lim_{M\rightarrow \infty} |\int_{0}^{M} f(x,y)dx| \rightarrow 0 $as $y \rightarrow 0$. It is given that $f(x,y)$ is bounded and measurable in $[0,\infty)\times [0,\infty).$
 A: No. Let $C\in \mathbb{R}_+$ and consider the bounded and measurable mapping $$f(x,y) = (y\land C)\times 1_{[0,\infty)}(x),$$
which obviously satisfies $f(x,y) \stackrel{y\to 0}{\to} 0$. As you stated it always hold that $|\int_0^M f(x,y) \, dx| \to 0$ as $y\to 0$ but we may also note that
$$
\left| \int_0^\infty f(x,y) \, dx \right|=(y\land C) \times \lambda([0,\infty))= \infty,
$$
for any $y\in(0,\infty)$, proving that the wanted convergence (interchanging of limits) is not possible. You might want read up on when one can interchange limits. Because if you can say that
$$
\lim_{y\to 0} \lim_{M\to \infty} \left| \int_0^M f(x,y) \, dx \right| = \lim_{M\to \infty}\lim_{y\to 0} \left| \int_0^M f(x,y) \, dx \right|,
$$
then your done.
Regarding your comment about uniform continuity:
No, assuming that you mean for any sequence of positive real numbers $(y_n)$ converging to zero, then the mappings $f_n(x):[0,\infty)\to\mathbb{R}$ given by $f_n(x)=f(x,y_n)$ converge uniformly to the zero mapping $x\mapsto 0$. In that case note that for any $C>\epsilon>0$ there exists an $N\in\mathbb{N}$ such that $|y_n-0| <  \epsilon $ for all $n\geq N$. Furthermore we have that
$$
|f_n(x)-0|=|(y_n \land C) 1_{[0,\infty)}(x)|<\epsilon
$$
for any $x\in[0,\infty)$. That is, we have just proved that our above example also converges uniformly to the zero function when $y\to 0$.
