Is Infimum of Non-Negative Measurable Functions Non-Negative Almost Everywhere?

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SUMMARY

The discussion centers on the proof that the infimum of a collection of non-negative measurable functions, {f_n}, defined on R, is non-negative almost everywhere. It is established that if |f_n(x)| ≤ 1 and f_n(x) ≥ 0 almost everywhere for all n in N, then f(x) = inf{f_n(x) | n in N} is also ≥ 0 almost everywhere on R. The proof utilizes the concept of sets with measure zero, concluding that the union of such sets does not affect the overall measure, thereby confirming the non-negativity of the infimum almost everywhere.

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Homework Statement


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.


Homework Equations



Almost everywhere, sequences of functions, measurability


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!
 
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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
 
Ah! Thanks for the input!
 

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