Showing a function $\notin L^1$

Question

Q) Let $F(x)=\frac{1}{x}\int_0^x f(t)dt$ for $x\in (0,\infty)$. If $f>0, f\in L^1$, show that $F\notin L^1$.

My attempt:

Since $f\in L^1$, $\int_0^x f(t)dt<\infty$ for all $x\in (0,\infty)$

$$\int_0^\infty |F(x)|dx = \int_0^\infty \frac{1}{x}\bigg|\int_0^x f(t)dt\bigg| = \int_0^\infty \frac{1}{x}\int_0^x f(t)dt \text{ since $f>0$ } \\ =\int_0^\infty \int_t^\infty \frac{f(t)}{x}dxdt \text{ By Fubini's } \\ =\int_0^\infty f(t)\log x\bigg|_t^\infty dt = \infty $$

Hence $F\notin L^1$. Is this correct?

Answer 1

Alternatively you have $F(x)\underset{+\infty}{\sim}\frac{\|f\|_1}{x}$ which is not integrable.