$f>0$ on $[0,1]$ implies $\int_0^1 f >0$

Solution 1:

Yes, your proof is correct. The two comments on that old answer are wrong. The first comment says:

The above proof is wrong: if $f(x)>C$ then $\int_0^1 f(x) \mathbb{d}x \geq C$. Notice that the inequality becomes non-strict, because integration is just passing to the limit and limits do not preserve the strictness of inequalities.

It is generally a good rule that limits do not preserve strict inequalities. But this user in his or her comment is wrong that this holds for integrals in particular. In fact, for the Lebesgue (or Riemann!) integral, if $f > g$ on a set $A$, and both are integrable, then $\int_A f > \int_A g$.

Specifically, let $A \subseteq \mathbb{R}$ be a Lebesgue measurable set, and suppose $f, g$ are measurable, with $f(x) > g(x)$ for all $x \in A$. Then $\int_A f > \int_A g$, with just a few exceptions:

• If $A$ has measure $0$, this won't hold.

• If $\int_A f = -\infty$ or if $\int_A g = \infty$, this won't hold.

This covers the Riemann case as well, since Riemann integrable functions are also Lebesgue integrable. (Except for some improper Riemann integrals -- I'm not sure if it holds in the case of improper integrals or not.)

This has also been covered on mathSE a lot of times. Some examples: 1, 2, 3, 4, 5.

Solution 2:

You are right.

Suppose $\int_0^1 f=0$.

Let $E_n=\{x \in [0,1] : f(x) > 1/n\}.$

$0=\int_0^1 f \geq \int_{E_n}f \geq \int_{E_n}(1/n)=m(E_n)(1/n)$

So $m(E_n)=0$ since $f>0 $ on $[0,1]$, $[0,1] = \cup E_n$ so $m([0,1])=0$.