Prove that $\int_0^1\int_x^1 \frac{f(y)}ydy\,dx=\int_0^1f(x)\,dx$ if $f$ is Lebesgue integrable

Question

Suppose f is a Lebesgue integrable function on [0,1] and define a new function by $$g(x)= \int_x^1 \frac{f(y)}y\text dy$$ for all x in [0,1]. Prove that

$$\int_0^1{g(x)}\text dx=\int_0^1{f(x)}\text dx$$

My approach: I thought of Fubini or Tonelli's theorem to use. But looks like that didn't work well. How do I approach this?

Answer 1

differentiating: $$ g'(x) = -\frac{f(x)}{x} $$ so $$ f(x) = - xg'(x) $$

integrating, then using parts for RHS $$ \int_0^1 f(x)\text dx = -\int_0^1 g'(x) x \text dx = [-xg(x)]_0^1 + \int_0^1 g(x) \text dx $$ and $$ [-xg(x)]_0^1 =0 $$ since $g(1)=0$

Answer 2

$$\int_0^1\int_x^1\frac{f(y)}{y}dydx=\int_0^1 \int_0^y\frac{f(y)}{y}dxdy=\int_0^1 \frac{f(y)}{y}\cdot ydy=\int_0^1 f(y)dy$$