# Show there exists a sequence of positive real numbers s.t. …

Let $f_n$ be a sequence of measurable functions on $[0,1]$ with $|f_n(x)|\lt\infty$ a.e. Show there exists a sequence $c_n$ of positive real numbers s.t. $f_n(x)/c_n\to0$ for almost every $x$ in $[0,1]$. Hint: use the borel-cantelli lemma- pick the sequence $c_n$ so that $$m[x{\rm\ in\ }[0,1]:|f_n(x)|/c_n\gt1/n]\lt2^{-n}$$ where $m$ is the measure.

• Presumably you mean to ask how to solve the (homework?) problem you posted and not order us to solve it. Please tell us what you have tried so far: e.g., you understand the relevance of the hint? – Pete L. Clark May 5 '13 at 3:06
• I don't understand how to use the hint. My first idea is to just have a sequence c sub n so that it is just infinity for each element in the sequence. Then surely the fn/cn will go to zero. I think of the hint as just a more difficult way of doing it. – Real Anal May 5 '13 at 3:20
• Perhaps the question should be reopened, as OP appears willing to engage in dialogue. – Gerry Myerson May 5 '13 at 7:04
• @RealAnal I fail to understand "so that it is just infinity for each element in the sequence". You might want to explain. – Did May 5 '13 at 7:12
• Unless there is a very good reason, we prefer not to arbitrarily delete good content: it could well help other users in the future. – robjohn May 13 '13 at 22:25

Fact 1: If $f:[0,1]\to\mathbb R$ is measurable and $|f(x)|$ is finite for almost every $x$, then for any $\epsilon>0$ there is a number $M$ such that the set $\{x:|f(x)|>M\}$ has measure at most $\epsilon$.
Reason: Since the intersection $\bigcap_{k=1}^\infty \{x:|f(x)|>k\}$ has measure $0$ and the sets in this intersection are nested, it follows that $m[\{x:|f(x)|>k\}]\to 0$ as $k\to\infty$.
Choose $c_n$ on the basis of Fact 1, so that $m[\{x:|f(x)|/c_n>1/n\}]<2^{-n}$.
The rest is an application of Borel-Cantelli: the set of number $x$ such that $|f(x)|/c_n>1/n$ infinitely often has measure zero. Rephrase this as: for almost every $x$ we have $|f(x)|/c_n\le 1/n$ for all sufficiently large $n$.