Why Does a Zero Integral Imply a Function Is Zero Almost Everywhere?

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If the integral of a non-negative measurable function f over a measure space (X,m) is zero, it implies that f must equal zero almost everywhere. The discussion revolves around proving that if f is positive on a set A with positive measure, then the integral over A would also be positive, contradicting the assumption that the integral over X is zero. The key argument is that if f is non-zero on a set of positive measure, the integral cannot be less than the integral over that set. Thus, the assumption leads to a contradiction, confirming that f must be zero almost everywhere. This conclusion reinforces the relationship between the integral of a function and its values across measurable sets.
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Homework Statement



Let (X,m) be a measure space, f:X->[0, infinity] be measurable.

If the integral over X of f is 0, show that f=0 almost everywhere



Homework Equations





The Attempt at a Solution



Suppose that f is nonzero on A, m(A)>0.

I've reduced problem to proving that the following is impossible:

f is positive on a set of nonzero measure and
{x such that f(x)>epsilon} has zero measure for all epsilon greater than zero.

I don't know how to do that though...

Any help?
 
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Suppose the statement is not true- that there exist a set A, of measure greater than 0, on which f is positive. Then the integral of f over A alone is positive and, since f is never negative, the integral of f over X is not less than the integral of f over A.
 
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