Bound for a integral

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Suppose $f(x)$ and $g(x)$ are two bounded nonnegative functions, does it hold that if $g(x)leq L$ then $$int_0^T f(x)g(x)dxleq L int_0^T f(x)dx$$



I think the answer is yes as it follows from monotonicity property of Lebesgue integral and $f(x)g(x)leq f(x)L$. Since $f(x)$ and $g(x)$ are bounded, the inequality above indeed make sense. Is it right?




real-analysis lebesgue-integral




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  • Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
    – James Yang
    Aug 2 at 12:30










  • @JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
    – mfl
    Aug 2 at 12:31










  • @JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
    – quallenjäger
    Aug 2 at 12:32











  • @mfl $g$ is bounded and thus integrable.
    – quallenjäger
    Aug 2 at 12:34










  • @JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
    – mfl
    Aug 2 at 12:37














up vote
3
down vote

favorite












Suppose $f(x)$ and $g(x)$ are two bounded nonnegative functions, does it hold that if $g(x)leq L$ then $$int_0^T f(x)g(x)dxleq L int_0^T f(x)dx$$



I think the answer is yes as it follows from monotonicity property of Lebesgue integral and $f(x)g(x)leq f(x)L$. Since $f(x)$ and $g(x)$ are bounded, the inequality above indeed make sense. Is it right?




real-analysis lebesgue-integral




share|cite|improve this question



















  • Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
    – James Yang
    Aug 2 at 12:30










  • @JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
    – mfl
    Aug 2 at 12:31










  • @JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
    – quallenjäger
    Aug 2 at 12:32











  • @mfl $g$ is bounded and thus integrable.
    – quallenjäger
    Aug 2 at 12:34










  • @JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
    – mfl
    Aug 2 at 12:37












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Suppose $f(x)$ and $g(x)$ are two bounded nonnegative functions, does it hold that if $g(x)leq L$ then $$int_0^T f(x)g(x)dxleq L int_0^T f(x)dx$$



I think the answer is yes as it follows from monotonicity property of Lebesgue integral and $f(x)g(x)leq f(x)L$. Since $f(x)$ and $g(x)$ are bounded, the inequality above indeed make sense. Is it right?




real-analysis lebesgue-integral




share|cite|improve this question











Suppose $f(x)$ and $g(x)$ are two bounded nonnegative functions, does it hold that if $g(x)leq L$ then $$int_0^T f(x)g(x)dxleq L int_0^T f(x)dx$$



I think the answer is yes as it follows from monotonicity property of Lebesgue integral and $f(x)g(x)leq f(x)L$. Since $f(x)$ and $g(x)$ are bounded, the inequality above indeed make sense. Is it right?





real-analysis lebesgue-integral





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 12:25









quallenjäger

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  • Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
    – James Yang
    Aug 2 at 12:30










  • @JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
    – mfl
    Aug 2 at 12:31










  • @JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
    – quallenjäger
    Aug 2 at 12:32











  • @mfl $g$ is bounded and thus integrable.
    – quallenjäger
    Aug 2 at 12:34










  • @JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
    – mfl
    Aug 2 at 12:37
















  • Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
    – James Yang
    Aug 2 at 12:30










  • @JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
    – mfl
    Aug 2 at 12:31










  • @JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
    – quallenjäger
    Aug 2 at 12:32











  • @mfl $g$ is bounded and thus integrable.
    – quallenjäger
    Aug 2 at 12:34










  • @JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
    – mfl
    Aug 2 at 12:37















Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
– James Yang
Aug 2 at 12:30




Yes. This is still true if f is just integrable. Nonnegativity is superfluous as well if you assume g is bounded.
– James Yang
Aug 2 at 12:30












@JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
– mfl
Aug 2 at 12:31




@JamesYang Is it not neccesary to assume that $g$ is also integrable? In other case $fg$ could not be integrable.
– mfl
Aug 2 at 12:31












@JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
– quallenjäger
Aug 2 at 12:32





@JamesYang Are you sure that nonnegativity is superfluous? As I cannot have $f(x)g(x) leq f(x)L$ almost everywhere if $f(x)$ is negative for some point.
– quallenjäger
Aug 2 at 12:32













@mfl $g$ is bounded and thus integrable.
– quallenjäger
Aug 2 at 12:34




@mfl $g$ is bounded and thus integrable.
– quallenjäger
Aug 2 at 12:34












@JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
– mfl
Aug 2 at 12:37




@JamesYang A bounded function is Lebesgue integrable if measurable. This is not assumed in the OP question.
– mfl
Aug 2 at 12:37















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