# Does showing a function is integrable suffice to show the function is measurable?

I am reviewing past homework exercises in preparation for a midterm exam. Fortunately, my professor provides solutions. However, I found one of his solutions contains an (seemingly) important argument that I didn't catch when I worked the exercise:

Let $f \in L^1[0,1]$. Set $g(x)=\int_x^1 \frac{f(t)}{t} dt$ for $0<x\leq 1$. Show that $g$ is defined a.e. on $[0,1]$, $g\in L^1[0,1]$, and $\int_0^1g(x)dx=\int_0^1f(x)dx$.

It is easy to see that $\int_x^1 \left|\frac{f(t)}{t}\right|<\infty$, hence $g(x)$ exists and finite for all $0<x\leq 1$. In my solution, I proceeded to find that $\int_0^1|g(x)|dx=\int_0^1\left|\int_x^1\frac{f(t)}{t}dt\right| dx$ is finite by means of Tonelli, and then that $\int_0^1g(x)dx=\int_0^1f(x)dx$ by means of Fubini.

In my professor's solution, however, after showing that $g(x)$ exists and finite for all $0<x\leq 1$, he shows $g$ is measurable! This makes sense because the Lebesgue integral is only defined for measurable functions (at least in our course).

My question arises here: did I inherently show $g$ is measurable by showing $\int_0^1|g(x)|dx=\int_0^1\left|\int_x^1\frac{f(t)}{t}dt\right| dx$ is finite, which I thought was a proof that $g\in L^1[0,1]$? Furthermore, my professor's proof to show $g$ is measurable is via showing $\int_0^1 \int_x^1 \left| \frac{f(t)}{t}\right|dtdx=\|f\|_1<\infty$. It seems he is claiming that showing $g$ to be integrable implies $g$ is measurable. But how can this be when the integral is only defined for measurable functions anyway?!?

Thanks in advance for any advice or clarification. I would also like any enlightening information about this function $g$.

• I realized now this is an application of Fubini's theorem. – Spanky Nov 7 '13 at 19:17