# Prove f>=0 almost everywhere

1. Dec 14, 2011

### jinsing

1. The problem statement, all variables and given/known data
If {f_n} is a collection of measurable functions defined on R and satisfying: |f_n(x)|<= 1 for all n in N and x in R and f_n(x) >=0 almost everywhere on R for all n in N and f(x) = inf{f_n(x)|n in N}, then f(x) >= 0 almost everywhere on R.

2. Relevant equations

Almost everywhere, sequences of functions, measurability

3. The attempt at a solution

Assume {f_n} is a collection of measurable functions defined on R satisfying the criteria above. Since for all n in N f_n(x) >= 0 a.e on R, define S_n = {x | f_n(x) < 0}. Then by definition for all n in N m(S_n) = 0. Now let $$S = \bigcup_{n=1}^\infty S_n.$$ Since S is the union of several sets S_n with measure of zero, then m(S) = 0. So, we know f(x) = inf{f_n(x) | n in N} >= 0 for all x in R-S. But m(S) = 0, so f(x) >= 0 almost everywhere, as desired.

I'm wondering if I can necessarily state that the inf of the f_n's will be >= 0. I'm not that great at catching little details like that, so having someone give the proof a quick look-over would be incredibly helpful.

Thanks!

2. Dec 14, 2011

### Office_Shredder

Staff Emeritus
It helps to clarify things if you fix x to be a specific value, say b. If $f_n(b) \geq$ for all values of n, then is $\inf\{f_n(b)} \geq 0$? This follows immediately from the definition of infimum.

Then conclude that S can only consist of values of x for which at least one fn(x)<0

3. Dec 15, 2011

### jinsing

Ah! Thanks for the input!