Proving that $g_nf_n$ converges to $0$ in measure on $[0,1]$ Question: Let $f_n,g_n:[0,1]\rightarrow [0,\infty)$ be measurable functions.  Assume $f_n\rightarrow 0$ in measure on $[0,1]$, and that $\int g_ndx<1$ for all $n\in\mathbb{N}$.  Prove that $g_nf_n\rightarrow 0$ in measure on $[0,1]$
My thoughts: So, we are trying to show that for all $\epsilon>0$, there exists $N$ such that $\forall n>N$, we have $m\{|g_nf_n-gf|>\epsilon\}<\epsilon$.  So, I was going to try and show convergence pointwise a.e. to then imply convergence in measure, but that fell apart.  So, would I go about this by setting up integrals and splitting the integral bounds?
Any help, suggestions, tips, etc. are (as always!) greatly appreciated!  Thank you.
 A: Since convergence in measure is equivalent the fact that every subsequence has  a further  subsequence converging almost everywhere we can reduce the proof to the case where $f_n$ tends to $0$ almost everywhere.
With this change the result can be proved easily using Egoroff's Theorem. Choose $E$ such that $f_n \to 0$ unifromly on $E$ and $\mu (E^{c}) <\epsilon$. Choose $n_0$ such that $ f_n(x) <\epsilon ^{2}$ for all $x \in E$ for all $n \geq n_0$. Now $\mu (f_ng_n >\epsilon) \leq \epsilon + \mu (E \cap (f_ng_n >\epsilon))\leq \epsilon+\mu (g_n >\frac 1 {\epsilon}) \leq \epsilon+\epsilon \int g_n d\mu<2\epsilon$.
A: $$
\begin{align}
\lambda(|f_ng_n|>\varepsilon)&=\lambda(|f_ng_n|>\varepsilon,|f_n|>\varepsilon^2)+\lambda(|f_ng_n|>\varepsilon,|f_n|\leq\varepsilon^2)\\
&\leq \lambda(|f_n|>\varepsilon^2) + \lambda(\varepsilon^2|g_n|>\varepsilon)\\
&\leq \lambda(|f_n|>\varepsilon^2) + \lambda(|g_n|>\tfrac{1}{\varepsilon})\leq 
\lambda(|f_n|>\varepsilon^2) + \varepsilon\int|g_n|\\
&\leq \lambda(|f_n|>\varepsilon^2) + \varepsilon
\end{align}
$$
Then for all $n$ large enough, say $n\geq N_\varepsilon$,
$$\lambda(|f_ng_n|>\varepsilon)\leq 2\varepsilon$$
This implies that $g_nf_n$ converges to $0$ in measure. To see this, let $\delta>0$. Choose $\varepsilon<\delta$ and let $N_\varepsilon$ be as above. Then
$\lambda(|f_ng_n|>\delta)\leq \lambda(|f_ng_n|>\varepsilon)\leq2\varepsilon$ for all $n\leq n\geq M_\varepsilon$. This shows that
$\lim_{n\rightarrow\infty}\lambda(|f_ng_n|>\delta)=0$.
