Lebesgue Integral Over Step Function

Given the step function

\begin{equation} g(x,y) = \begin{cases} \frac{1}{x^2} & 0<y<x<1\\ -\frac{1}{y^2} & 0<x<y<1\\ 0 & \text{otherwise} \end{cases} \end{equation}

I want to show that

$$\int_{(0,1)}\int_{(0,1)} g(x,y) \, d\lambda(x)\,d\lambda(y) \neq \int_{(0,1)}\int_{(0,1)} g(x,y) \,d\lambda(y)\,d\lambda(x)$$

However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as

$$\int_0^x \int_0^1 \frac{1}{x^2} \, dx \, dy - \int_0^1 \int_0^y \frac{1}{y^2} \, dx \, dy$$

I also considered using limits such as following one, but again I end up with divergence.
$$\lim_{k \rightarrow\infty} \int_{(\frac{1}{k},1)}\lim_{k \rightarrow\infty} \int_{(\frac{1}{k},1)}g(x,y) \, dx \, dy$$

I would really appreciate any help!

Solution 1:

\begin{align} & \int_{(0,1)} \left( \int_{(0,1)} g(x,y) \, d\lambda(x) \right) \,d\lambda(y) \\[8pt] = {} & \int_{(0,1)} \left( \int_{(0,y)} -\frac 1 {y^2} \, d\lambda(x) + \int_{(y,1)} \frac 1 {x^2} \,d\lambda(x) \right) \, d\lambda(y) \\[8pt] = {} & \int_{(0,1)} \left( -\frac 1 y + \frac{1-y} y \right) \, d\lambda(y) =\int_{(0,1)} -1\, d\lambda(y) = -1. \end{align}