One application of Egoroff's Theorem There is such a question:
Denote Lebesgue measure by $\lambda$. Let $(f_k)_{k=1}^\infty$ be a sequence of nonnegative Lebesgue integrable functions on $[0,1]$. Assume that $\int_{[0,1]}f_k \,d\lambda=1$ and that there is a constant $M<\infty$ so that $f_k(t)\le M$ for all $k$ and all $t\in [0,1]$. If $(a_k)_{k=1}^\infty$ is a nonnegative real sequence such that $\sum_{k=1}^\infty a_k=\infty$, show that there is a Lebesgue measurable subset $E$ of $[0,1]$ with $\lambda(E)>0$ such that $\sum_{k=1}^\infty a_k f_k(t)=\infty$ for all $t\in E$.
The teacher gives us hints to try to use the Egoroff's theorem, however, I don't have any ideals about dealing with $\{a_k\}$. Wish you could help me.
${{}}$
 A: Assume by contradiction that the series converges for a.e. $t\in [0,1]$.
Since $a_k\geq 0$ and $f_k\geq 0$, this is equivalent to assume that the sequence
$$
F_n(t) := \sum_{k=1}^n a_k f_k(t)
$$
converges pointwise to $F(t) := \sum_{k=1}^\infty a_k f_k(t) < +\infty$ for a.e. $t\in [0,1]$.
Let us fix $\varepsilon>0$, $\varepsilon < 1/(2M)$. By Egorov's theorem, there exists a measureble set $A_\varepsilon \subset [0,1]$ such that
$$
\lambda([0,1]\setminus A_\varepsilon) < \varepsilon,
\qquad
F_n \to F\ \text{uniformly in}\ A_\varepsilon.
$$
But then $F$ is integrable in $A_\varepsilon$ and
$$
\int_{A_\varepsilon} F \,d\lambda = \lim_n \int_{A_\varepsilon} F_n \,d\lambda
= \sum_{k=1}^\infty a_k \int_{A_\varepsilon} f_k \,d\lambda 
=: \sum_{k=1}^\infty a_k I_k < +\infty
$$
On the other hand, $\int_0^1 f_k \,d\lambda = 1$ and $0\leq f_k\leq M$, so that
$$
1-I_k = \int_{[0,1]\setminus A_\varepsilon} f_k \,d\lambda \leq M\varepsilon,
$$ 
so that $I_k \geq 1 - M\varepsilon \geq 1/2$ for every $k$, so that $\sum_k a_k I_k$ diverges to $+\infty$, a contradiction.
A: By contradiction: Assume $\sum_ka_kf_k(t)<+\infty$ except on a set of zero measure $Z$. Then, for $t\notin Z$, $f_k(t)\to 0$.(*) Ie, $f_k$ converges pointwise to $0$.
By Egoroff, for all $\epsilon>0$, there is some $B\subset [0,1]$, with $\lambda(B)<\epsilon$, s.t. $f_k$ converges unformly to $0$ on $[0,1]\setminus B$. Taking $\epsilon = \frac{1}{2M}$, there is some $B\subset [0,1]$, with $\lambda(B)<\frac{1}{2M}$ s.t. $f_k$ converges unformly to $0$ on $[0,1]\setminus B$. 
Hence, for all $f_n$, 
\begin{equation}
\int_Bf_nd\lambda < M\cdot \frac{1}{2M} =\frac{1}{2}
\end{equation}
Also, by uniform continuity outside $B$, for all $\epsilon>0$, there is $n_0$ s.t. $n\ge n_0\implies ||f_n|_{[0,1]\setminus B}||<\epsilon$. Hence, taking $\epsilon = \frac{1}{2}$, and large enough $n\ge n_0$, we get
\begin{equation}
\int_{[0,1]\setminus B}f_nd\lambda < \epsilon = \frac{1}{2}
\end{equation}
Hence, we conclude that for large enough $n$, $\int_{[0,1]}f_nd\lambda < 1$, a contradiction.
Proof of (*): Otherwise, for some such $t\notin Z$, there is some $\epsilon>0$, such that $f_k(t)>\epsilon$. Hence, for all $n$, $\sum_{k=1}^na_kf_k(t)\ge\epsilon\sum_{k=1}^na_k$ which diverges to infinity. Which contradicts $\sum_ka_kf_k(t)<+\infty$
