If $\int_0^1 f_n dx \leq a_n$ where $\sum_{n=1}^{\infty}a_n < \infty$ show $f_n \to 0$

Suppose $\{f_n\}_{n=1}^{\infty}$ is a sequence of nonnegative measurable functions on $[0,1]$ with $$\int_0^1 f_n \, dx \leq a_n \qquad \text{for all} \quad n\geq 1$$ where $$\sum_{n=1}^{\infty}a_n < \infty.$$ Prove that $f_n \to 0$ a.e on $[0,1]$

What I am able to do so far

$f_n\geq 0$, measurable , $x\in [0,1]$

$f_n \geq 0 \Rightarrow 0 \leq \int_0^1 f_ndx \leq a_n$ , $\Rightarrow a_n \geq 0, \forall n$

Hence $\{a_n\}$ is convergent. thus $\sum_{n=1}^{\infty}|a_n| <\infty$ Therefore $\forall \epsilon >0$, $\exists N$such that $\forall n\geq N$ $$\sum_{n=N}^{\infty}a_n<\epsilon$$ and so $$n\geq N, 0\leq \int_0^1 f_n \leq \epsilon$$ But $\epsilon$ is arbitrary $\Rightarrow \int_0^1 f_n \to 0$

That is to say the integral converges to $0$, but how do I show that the sequence itself converges to $0$

• I'd use the Fubini-Tonelli theorem. – tristan Sep 8 '17 at 18:29

Hint: $$\sum_{n=1}^{\infty}\int_0^1f_n dx \lt \infty\implies \int_0^1\left(\sum_{n=1}^{\infty}f_n\right)dx \lt \infty$$

This shows that $$\sum_{n=1}^{\infty}f_n \lt \infty \text { -a.e}$$

• Thank you, and sorry for the delay in response, the Hurricane got me running. If I understand the hint, I have $\int_0^1f_n \to 0$ and $\sum_{n=1}^{\infty}f_n <\infty$ can I conclude that since $f_n\geq 0$, and the sum converges it implies that $\forall \epsilon > 0$ ,$\exists N$ such that $\forall n\geq N$ ,$f_n<\epsilon$ – Cnine Sep 11 '17 at 17:21
• Basically except possibly a null set, the sum converges and hence the $n$-th term goes to $0$. Hope you are safe!! – tattwamasi amrutam Sep 11 '17 at 17:25
• Oh okay, very helpful. thank you very much. And yes I am safe :) – Cnine Sep 11 '17 at 17:29

Convergence of the integrals $\int_0^1 f_n \to 0$ does, in general, not imply $f_n \to 0$; therefore your approach doesn't work.

Hints:

1. Recall the Borel-Cantelli lemma: Let $\mu$ be a finite measure. If $(A_n)_{n \in \mathbb{N}}$ is a sequence of measurable sets such that $$\sum_{n \geq 1} \mu(A_n)<\infty$$ then $$\mu \left( \limsup_{n \to \infty} A_n \right) = 0.$$
2. Define $$A_n := \{x \in [0,1]; |f_n(x)| \geq 1/k\}$$ for fixed $k \geq 1$. Use Markov's inequality and the Borel-Cantelli lemma to show that $\limsup_{n \to \infty} A_n$ has Lebesgue measure $0$.
3. By step 2 there exists for any $k \in \mathbb{N}$ a null set $N_k$ such that for any $x \in [0,1] \backslash A_k$ it holds that $$|f_n(x)| \leq \frac{1}{k}$$ for $n \gg 1$ sufficiently large. Conclude that $f_n(x) \to 0$ for any $x \in [0,1] \backslash \left( \bigcup_{k \geq 1} N_k \right)$.

By the monotone convergence theorem, $$\int_0^1\sum_{n=1}^\infty f_n = \sum_{n=1}^\infty\int_0^1 f_n \le \sum_{n=1}^\infty \int_0^1a_n =\sum a_n< \infty.$$ Thus, for almost every $x$ we have that $\sum_{n=1}^\infty f_n(x) < \infty$. Since the series $\sum f_n(x)$ is convergent for almost every $x$, we can conclude that $f_n(x)\to 0$ for almost every $x$.