Convergence of a series given properties on uniformly bounded functions I have the following problem from "Bollobas, Linear analysis. An introductory course"
Let $\phi_n:[0,1]\to\mathbb{R}^+$ $(n=1,2,\dots)$ be uniformly bounded continuous function such that $$\int_{0}^1\phi_n(x)dx\geq c$$ for some $c>0$. Suppose $c_n\geq 0$, $(n=1,2,\dots)$ and $$\sum_{n\geq 1}c_n\phi_n(x)<\infty$$ for every $x\in[0,1]$. Prove that $$\sum_{n\geq 1}c_n<\infty$$
I don't know how to proceed. Any hints will be useful !
Thanks in advance
 A: We know that the sequence $(\varphi_n)_{n \in \mathbb{N}}$ is uniformly bounded, i.e. we have $|\varphi_n|_\infty \le M$ for all $n \in \mathbb{N}$ for some $M>0$ independent of the index $n$. Moreover, we write $B_n:= \{\varphi_n \ne 0 \}$ and let $$X := \bigcup_{n=1}^\infty B_n.$$ Note that $X$ is open and not empty. In fact, we have $B_n \neq \emptyset$ for all $n \in \mathbb{N}$, because of $\int_0^1 \varphi_n(t) \, dt \ge c$. In particular, $0< \lambda(X) \le 1$. (Moreover, $X$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define
$$h(x) := \sum_{k=1}^\infty c_k \varphi_k(x).$$
This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := \{x \in X : n < h(x) \le n+1\}$$ and note that $X = \bigcup_{n=1}^\infty A_n$.
By $\sigma$-additivity  of the measure, we have
$$ \lambda(X) = \sum_{k=1}^\infty \lambda(A_k).$$
Hence the series is convegent and we find for some $N \in \mathbb{N}$ that
$$\sum_{k=N}^\infty \lambda(A_k) < \frac{c}{2M}.$$
Set $Y:= \bigcup_{k=1}^N A_k$. Note that for any $x \in Y$ we have $|h(x)| \le N+1$ and $\lambda(X \setminus Y) < c/(2M)$. Therefore we get
$$\int_Y \varphi_n(x) \,dx \ge c- \int_{X \setminus Y} \varphi_n(x) \,dx \ge \frac{c}{2}.$$
The monotone convergence theorem implies now
$$\frac{c}{2} \sum_{n=1}^\infty c_n \le \sum_{k=1}^\infty c_k  \int_Y \varphi_k(x) \, dx  = \int_Y \sum_{k=1}^\infty c_k \varphi_k(x) \, dx = \int_Y h(x) \, dx \le N+1.$$
Thus the series in the last equation is convergent.
