Let $\{f_n\}$ be a sequence of $\textbf{continuous}$ real valued functions defined on $[0, \infty)$. Suppose $f_n(x) \rightarrow f(x)$ for all $x \in [0,\infty)$ and that $f$ is integrable. Then
a) $\int_0^\infty f_n(x)\;dx \rightarrow \int_0^\infty f(x)\; dx$ as $n \rightarrow \infty$
b) If $f_n \rightarrow f$ uniformly on $[0,\infty)$, then $\int_0^1f_n(x)\;dx \rightarrow \int_0^1 f(x)\; dx$
c) If $f_n \rightarrow f$ uniformly on $[0,\infty)$, then $\int_0^\infty f_n(x)\;dx \rightarrow \int_0^\infty f(x)\; dx$
d) If $\int_0^1 \vert f_n(x- f(x) \vert dx \rightarrow 0$, then$f_n \rightarrow f$ uniformly on $[0,1]$
b) is true, since , if $f_n \rightarrow f$ uniformly on $[0,\infty)$, then it is also uniformly convergent in $[0,1]$, so by a familiar result on integration, $f$ is continuous and $\lim_{n \rightarrow \infty} \int_0^1 f_n(x)\; dx=\int_0^1 f(x)\;dx$
What about others? Any hint must be appreciated!
Let $f_n(x)=\frac 1 {n(1+|x|)}$ and $f(x)=0$. Then $f_n \to f$ uniformly but $\int f_n(x) \, dx =\infty$ for all $n$. Hence a) and c) are false.
d) is also false:
Let
$$f_n(x) = \begin{cases} nx&(0 \leq x \leq \frac{1}{n})\\ 1 &(x> \frac{1}{n})\\ \end{cases}$$
Let $f(x)=1$ for all $x$. Then $$\int_0^{1} |f_n(x)-f(x)|\, dx=\frac {1} {2n} \to 0$$ but $$\sup \{|f_n(x)-f(x)|:x \in [0,1]\} \geq |f_n(0)-f(0)|=1$$ for all $n$, so $f_n$ does not tend to $f$ uniformly.