I have the following task:
Comparison Lemma. Suppose $-\infty \leq a \lt b \leq \infty$ and that $f,F$ are continuous on $(a,b)$. If the improper integrl $\int^b_aF(x)dx$ converges and $|f(x)| \leq F(x) , \forall x \in (a,b)$, then the improper integral $\int^b_af(x)dx$ converges as well.
Prove the test for the case $a=0,b=1$.
Here's how I try to prove it:
If $\int^1_0F(x)dx$ converges then $\int^1_0F(x)dx=A$ for some $A \in \Bbb R_{+}$ and since $$|f(x)| \leq F(x) \Leftrightarrow \int^1_0|f(x)|dx \leq \int^1_0F(x)dx=A\Leftrightarrow \int^1_0|f(x)|dx\leq A\Leftrightarrow|\int^1_0f(x)dx|\leq\int^1_0|f(x)|dx\leq A\Leftrightarrow -A\leq\int^1_0f(x)dx\leq A$$
And hence $\int^1_0f(x)dx$ converges
Is my proof correct?
You just say that if $\int_0^1|f|$ converges, then $\int_0^1|f|\leq A$. Not that $\int_0^1|f|$ converges.
Hint
Remark that if $y\in (a,b)$, then $$x\mapsto \int_y^x|f|,$$ in increasing and upper bounded. Also, $$y\mapsto \int_y^1|f|$$ is decreasing and lower bounded.