Let $f$ be Lebesgue integrable on $(0, 1)$. For $0 < x < 1$ define g(x) = $\int_x^1t^{-1}f(t)dt$
Prove that $g$ is Lebesgue integrable on $(0, 1)$.
$\int^1_0g(x)dx=\int^1_0f(x)dx.$
I am not really getting idea from where should i start. I tried supposing f(x) as characteristics function then simple function and approximating f(x) by the simple function. I don't think this is the right idea. Could anyone give me some hint how to start up.
Okay, let me give you the full version of the Fubini-solution hinted at in the comments:
By the Fubini-Tonelli theorem (for non-negative measurable functions also applicable without integrability assumptions) $$ \begin{align} & \phantom{=}\int_{0}^{1}\int_{x}^{1}t^{-1}\left|f\left(t\right)\right|\, dt\, dx \\ & \overset{\left(\ast\right)}{=}\int_{0}^{1}\int_{0}^{t}t^{-1}\left|f\left(t\right)\right|\, dx\, dt\\&=\int_{0}^{1}t^{-1}\left|f\left(t\right)\right|\cdot\int_{0}^{t} 1\, dx\, dt\\&=\int_{0}^{1}\left|f\left(t\right)\right|\, dt<\infty. \end{align} $$ Here, the step marked with $\left(\ast\right)$ used the equivalence $t\geq x\,\Leftrightarrow\, x\leq t$, you could make this more precise using $$ \int_{0}^{1}\int_{x}^{1}h\left(t\right)\, dt\, dx=\int_{0}^{1}\int_{0}^{1}\chi_{\left\{ \left(a,b\right) \mid 0\leq a\leq b\leq1\right\} }\left(x,t\right)\cdot h\left(t\right)\, dt\, dx, $$ where $\chi_M$ is the characteristic function of $M$.
As the above integral (with the absolute values) is finite, you can then apply Fubini in just the same way without the absolute value around $f$ to conclude the proof.