Integral of product measure

Let $f_k= k^{-2}2^{-k}1_{[-k, k)\times [-k,k)}$. Then 
\[\int_{\R^2} f_k d\lambda^2= \int_{\R^2}k^{-2}2^{-k}1_{[-k, k)\times [-k,k)}d\lambda^2\]
\[=k^{-2}2^{-k} \int_{\R^2}1_{[-k, k) \int_{\R^2}\times [-k,k)}d\lambda^2=k^{-2}2^{-k} \times (2k \times 2k)=2^{2-k}.\]

On the other hand, $|f_k(x)|\leq k^{-2}2^{-k}$, and $\sum_{k=1}^{\infty}k^{-2}2^{-k}$ is convergent. Hence by Weierstrass M-test (https://en.wikipedia.org/wiki/Weierstrass_M-test)
\[g_n= \sum_{k=1}^{n} f_k\]
converges uniformly on $\R^2$, therefore we can exchange the order of the limit and integration. Thus 
\[\int_{\R^2} \sum_{k=1}^{\infty}f_k d\lambda^2=\int_{\R^2}\lim_{n \rightarrow \infty}\sum_{k=1}^{n}f_k d\lambda^2\]
\[\lim_{n \rightarrow \infty}\int_{\R^2}\sum_{k=1}^{n}f_k d\lambda^2= \lim_{n \rightarrow \infty}\sum_{k=1}^{n}\int_{\R^2}f_k d\lambda^2\]
\[=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}2^{2-k}=4.\]