Proving that $g_nf_n$ converges to $0$ in measure on $[0,1]$

$$ begin{align} lambda(|f_ng_n|>varepsilon)&=lambda(|f_ng_n|>varepsilon,|f_n|>varepsilon^2)+lambda(|f_ng_n|>varepsilon,|f_n|leqvarepsilon^2)\ &leq lambda(|f_n|>varepsilon^2) + lambda(varepsilon^2|g_n|>varepsilon)\ &leq lambda(|f_n|>varepsilon^2) + lambda(|g_n|>tfrac{1}{varepsilon})leq lambda(|f_n|>varepsilon^2) + varepsilonint|g_n|\ &leq lambda(|f_n|>varepsilon^2) + varepsilon end{align} $$

Then for all $n$ large enough, say $ngeq N_varepsilon$, $$lambda(|f_ng_n|>varepsilon)leq 2varepsilon$$ 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 $lambda(|f_ng_n|>delta)leq lambda(|f_ng_n|>varepsilon)leq2varepsilon$ for all $nleq ngeq M_varepsilon$. This shows that $lim_{nrightarrowinfty}lambda(|f_ng_n|>delta)=0$.

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 $mu (f_ng_n >epsilon) leq epsilon + mu (E cap (f_ng_n >epsilon))leq epsilon+mu (g_n >frac 1 {epsilon}) leq epsilon+epsilon int g_n dmu<2epsilon$.