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Gerald Edgar
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The condition $$ \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$$$ \sum_{n=1}^\infty \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$ will let you do 1,2. Alternatively, $f_n(u,v,w,t)\ge 0$ will also let you do 1,2 wit the proviso that you have to allow value $+\infty$ for both sides.

The condition $$ \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$ will let you do 1,2. Alternatively, $f_n(u,v,w,t)\ge 0$ will also let you do 1,2 wit the proviso that you have to allow value $+\infty$ for both sides.

The condition $$ \sum_{n=1}^\infty \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$ will let you do 1,2. Alternatively, $f_n(u,v,w,t)\ge 0$ will also let you do 1,2 wit the proviso that you have to allow value $+\infty$ for both sides.

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Gerald Edgar
  • 41.1k
  • 5
  • 125
  • 219

The condition $$ \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$ will let you do 1,2. Alternatively, $f_n(u,v,w,t)\ge 0$ will also let you do 1,2 wit the proviso that you have to allow value $+\infty$ for both sides.